NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

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# NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

Edited By Ramraj Saini | Updated on Sep 25, 2023 07:13 PM IST

## Conic Sections Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections are provided here. As the name suggests a conic section is a curve obtained from the intersection of the surface of a cone with a plane. There are three types of conic section hyperbola, the parabola, and the ellipse discussed in the Class 11 NCERT syllabus. The circle is a special case of the ellipse which has been discussed in this class 11 maths NCERT book chapter. In the conic sections class 11 questions and answers, you will see problems related to above-mentioned curves like circles, parabolas, hyperbolas and ellipses. These are developed by expert team of Careers360 based on the latest syllabus of CSBE 2023. In the class 11 maths chapter 11 question answer, you will get solutions to miscellaneous exercise too. Here students can find NCERT solutions for class 11 at single place.

The CBSE syllabus features of conic sections class 11, which covers key topics including Conic Sections, Sections of a Circle, as well as Circle, Parabola, Hyperbola, and Ellipse. these concepts are essential for both CBSE exam and competitive exams like JEE Mains, JEE Advanced, VITEEE, BITSAT etc because every year many questions are asked from this topic. There is a total of 62 questions are given in 4 exercises of NCERT textbook. All these questions are explained in the class 11 maths chapter 11 NCERT solutions. These curves have applications in fields like the design of antennas and telescopes, planetary motion, reflectors in automobile headlights, etc. There are 8 questions in a miscellaneous exercise.

## Conic Sections Class 11 Solutions - Important Formulae

Circle:

 Description Equation/Formulas Equation of a circle (x - h)2 + (y - k)2 = r2 General equation of a circle x2 + y2 + 2gx + 2fy + c = 0 Center of the circle Centre: (-g, -f) Radius of the circle Radius (r) = √(g2 + f2 - c) Parametric equation of a circle x = r cos(θ), y = r sin(θ) Parametric equation of a circle (centred at h, k) x = h + r cos(θ), y = k + r sin(θ)

Parabola:

 Description Equations/Forms Equation forms of parabola y2 = 4ax, y2 = -4ax, x2 = 4ay, x2 = -4ay Axis of the parabola y = 0 (for first two forms), x = 0 (for last two forms) Directrix of the parabola x = -a (1st form), x = a (2nd form), y = -a (3rd form), y = a (4th form) Vertex of the parabola (0, 0) (for all forms) Focus of the parabola (a, 0) (1st form), (-a, 0) (2nd form), (0, a) (3rd form), (0, -a) (4th form) Length of latus rectum 4a (for all forms) Focal length `

Ellipse:

 Description Equation/Forms Equation forms of ellipse x2/a2 + y2/b2 = 1 (a > b), x2/b2 + y2/a2 = 1 (a > b) Major Axis y = 0 (1st form), x = 0 (2nd form) Length of Major Axis 2a (for both forms) Minor Axis x = 0 (1st form), y = 0 (2nd form) Length of Minor Axis 2b (for both forms) Directrix of the ellipse x = ±a/e (1st form), y = ±a/e (2nd form) Vertex of the ellipse (±a, 0) (1st form), (0, ±a) (2nd form) Focus of the ellipse (±ae, 0) (1st form), (0, ±ae) (2nd form) Length of latus rectum 2b2/a (for both forms) Eccentricity (e) √(a2 + b2)/a2 (for both forms)

conic sections class 11 questions and answers

 Description Equations/Forms Equation forms of hyperbola x2/a2 - y2/b2 = 1, x2/a2 - y2/b2 = -1 Coordinates of centre (0, 0) (for both forms) Coordinates of vertices (±a, 0) (for both forms) Coordinates of foci (±ae, 0) (for both forms) Length of Conjugate axis 2b (for both forms) Length of Transverse axis 2a (for both forms) Equation of Conjugate axis x = 0 (for both forms) Equation of Transverse axis y = 0 (for both forms) Equation of Directrix x = ±a/e (for both forms) Eccentricity (e) √(a2 + b2)/a2 (for both forms) Length of latus rectum 2b2/a (for both forms)

Free download NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections for CBSE Exam.

## Conic Sections Class 11 NCERT Solutions (Intext Questions and Exercise)

NCERT conic sections class 11 solutions - Exercise: 11.1

Question:1 Find the equation of the circle with

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=(0,2)$

AND $r=2$

So the equation of the circle is:

, $(x-0)^2+(y-2)^2=2^2$

$x^2+y^2-4y+4=4$

$x^2+y^2-4y=0$

Question:2 Find the equation of the circle with

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=(-2,3)$

AND $r=4$

So the equation of the circle is:

, $(x-(-2))^2+(y-3)^2=4^2$

$x^2+4x+4+y^2-6y+9=16$

$x^2+y^2+4x-6y-3=0$

Question:3 Find the equation of the circle with

As we know,

The equation of the circle with center ( h, k) and radius r is give by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=\left ( \frac{1}{2},\frac{1}{4} \right )$

AND

$r=\frac{1}{12}$

So the equation of circle is:

, $\left ( x-\frac{1}{2}\right )^2+\left ( y-\frac{1}{4}\right )^2=\left ( \frac{1}{12}\right )^2$

$x^2-x+\frac{1}{4}+y^2-\frac{1}{2}y+\frac{1}{16}=\frac{1}{144}$

$x^2+y^2-x-\frac{1}{2}y-\frac{11}{36}=0$

$36x^2+36y^2-36x-18y-11=0$

Question:4 Find the equation of the circle with

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=(1,1)$

AND $r=\sqrt{2}$

So the equation of the circle is:

, $(x-1)^2+(y-1)^2=(\sqrt{2})^2$

$x^2-2x+1+y^2-2y+1=2$

$x^2+y^2-2x-2y=0$

Question:5 Find the equation of the circle with

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=(-a,-b)$

AND $r=\sqrt{a^2-b^2}$

So the equation of the circle is:

, $(x-(-a))^2+(y-(-b))^2=(\sqrt{a^2-b^2})^2$

$x^2+2ax+a^2+y^2+2by+b^2=a^2-b^2$

$x^2+y^2+2ax+2by+2b^2=0$

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given here

$(x+5)^2 + (y-3)^2 = 36$

Can also be written in the form

$(x-(-5))^2 + (y-3)^2 = 6^2$

So, from comparing, we can see that

$r=6$

Hence the Radius of the circle is 6.

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given here

$x^2 + y^2 -4x - 8y - 45 = 0$

Can also be written in the form

$(x-2)^2 + (y-4)^2 =(\sqrt{65})^2$

So, from comparing, we can see that

$r=\sqrt{65}$

Hence the Radius of the circle is $\sqrt{65}$ .

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given here

$x^2 + y^2 -8x +10y -12 = 0$

Can also be written in the form

$(x-4)^2 + (y-(-5))^2 = (\sqrt{53})^2$

So, from comparing, we can see that

$r=\sqrt{53}$

Hence the radius of the circle is $\sqrt{53}$ .

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given here

$2x^2 + 2y^2 - x = 0$

Can also be written in the form

$\left ( x-\frac{1}{4}\right )^2 + \left ( y-0 \right )^2 = \left ( \frac{1}{4} \right )^2$

So, from comparing, we can see that

$r=\frac{1}{4}$

Hence Center of the circle is the $\left ( \frac{1}{4},0\right )$ Radius of the circle is $\frac{1}{4}$ .

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given Here,

Condition 1: the circle passes through points (4,1) and (6,5)

$(4-h)^2+(1-k)^2=r^2$

$(6-h)^2+(5-k)^2=r^2$

Here,

$(4-h)^2+(1-k)^2=(6-h)^2+(5-k)^2$

$(4-h)^2-(6-h)^2+(1-k)^2-(5-k)^2=0$

$(-2)(10-2h)+(-4)(6-2k)=0$

$-20+4h-24+8k=0$

$4h+8k=44$

Now, Condition 2:centre is on the line $4x + y = 16$ .

$4h+k=16$

From condition 1 and condition 2

$h=3,\:k=4$

Now lets substitute this value of h and k in condition 1 to find out r

$(4-3)^2+(1-4)^2=r^2$

$1+9=r^2$

$r=\sqrt{10}$

So now, the Final Equation of the circle is

$(x-3)^2+(y-4)^2=(\sqrt{10})^2$

$x^2-6x+9+y^2-8y+16=10$

$x^2+y^2-6x-8y+15=0$

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given Here,

Condition 1: the circle passes through points (2,3) and (–1,1)

$(2-h)^2+(3-k)^2=r^2$

$(-1-h)^2+(1-k)^2=r^2$

Here,

$(2-h)^2+(3-k)^2=(-1-h)^2+(1-k)^2$

$(2-h)^2-(-1-h)^2+(3-k)^2-(1-k)^2=0$

$(3)(1-2h)+(2)(4-2k)=0$

$3-6h+8-4k=0$

$6h+4k=11$

Now, Condition 2: centre is on the line. $x - 3y - 11 = 0$

$h-3k=11$

From condition 1 and condition 2

$h=\frac{7}{2},\:k=\frac{-5}{2}$

Now let's substitute this value of h and k in condition 1 to find out $r$

$\left ( 2-\frac{7}{2}\right )^2+\left (3+\frac{5}{2}\right )^2=r^2$

$\frac{9}{4}+\frac{121}{4}=r^2$

$r^2=\frac{130}{4}$

So now, the Final Equation of the circle is

$\left(x-\frac{7}{2}\right )^2+\left(y+\frac{5}{2}\right)^2=\frac{130}{4}$

$x^2-7x+\frac{49}{4}+y^2+5y+\frac{25}{4}=\frac{130}{4}$

$x^2+y^2-7x+5y-\frac{56}{4}=0$

$x^2+y^2-7x+5y-14=0$

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So let the circle be,

$(x-h)^2+(y-k)^2=r^2$

Since it's radius is 5 and its centre lies on x-axis,

$(x-h)^2+(y-0)^2=5^2$

And Since it passes through the point (2,3).

$(2-h)^2+(3-0)^2=5^2$

$(2-h)^2=25-9$

$(2-h)^2=16$

$(2-h)=4\:or\:(2-h)=-4$

$h=-2\: or\;6$

When $h=-2\:$ ,The equation of the circle is :

$(x-(-2))^2+(y-0)^2=5^2$

$x^2+4x+4+y^2=25$

$x^2+y^2+4x-21=0$

When $h=6$ The equation of the circle is :

$(x-6)^2+(y-0)^2=5^2$

$x^2-12x+36+y^2=25$

$x^2+y^2-12x+11=0$

Let the equation of circle be,

$(x-h)^2+(y-k)^2=r^2$

Now since this circle passes through (0,0)

$(0-h)^2+(0-k)^2=r^2$

$h^2+k^2=r^2$

Now, this circle makes an intercept of a and b on the coordinate axes.it means circle passes through the point (a,0) and (0,b)

So,

$(a-h)^2+(0-k)^2=r^2$

$a^2-2ah+h^2+k^2=r^2$

$a^2-2ah=0$

$a(a-2h)=0$

$a=0\:or\:a-2h=0$

Since $a\neq0\:so\:a-2h=0$

$h=\frac{a}{2}$

Similarly,

$(0-h)^2+(b-k)^2=r^2$

$h^2+b^2-2bk+k^2=r^2$

$b^2-2bk=0$

$b(b-2k)=0$

Since b is not equal to zero.

$k=\frac{b}{2}$

So Final equation of the Circle ;

$\left ( x-\frac{a}{2} \right )^2+\left ( y-\frac{b}{}2 \right )^2=\left ( \frac{a}{2} \right )^2+\left ( \frac{b}{2} \right )^2$

$x^2-ax+\frac{a^2}{4}+y^2-bx+\frac{b^2}{4}=\frac{a^2}{4}+\frac{b^2}{4}$

$x^2+y^2-ax-bx=0$

Let the equation of circle be :

$(x-h)^2+(y-k)^2=r^2$

Now, since the centre of the circle is (2,2), our equation becomes

$(x-2)^2+(y-2)^2=r^2$

Now, Since this passes through the point (4,5)

$(4-2)^2+(5-2)^2=r^2$

$4+9=r^2$

$r^2=13$

Hence The Final equation of the circle becomes

$(x-2)^2+(y-2)^2=13$

$x^2-4x+4+y^2-4y+4=13$

$x^2+y^2-4x-4y-5=0$

Given, a circle

$x^2 + y^2 = 25$

As we can see center of the circle is ( 0,0)

Now the distance between (0,0) and (–2.5, 3.5) is

$d=\sqrt{(-2.5-0)^2+(3.5-0)^2}$

$d=\sqrt{6.25+12.25}$

$d=\sqrt{18.5}\approx 4.3$ $d=\sqrt{18.5}\approx 4.3<5$

Since distance between the given point and center of the circle is less than radius of the circle, the point lie inside the circle.

Conic sections class 11 NCERT solutions - Exercise: 11.2

Given, a parabola with equation

$y^2 =12x$

This is parabola of the form $y^2=4ax$ which opens towards the right.

So,

By comparing the given parabola equation with the standard equation, we get,

$4a=12$

$a=3$

Hence,

Coordinates of the focus :

$(a,0)=(3,0)$

Axis of the parabola:

It can be seen that the axis of this parabola is X-Axis.

The equation of the directrix

$x=-a,\Rightarrow x=-3\Rightarrow x+3=0$

The length of the latus rectum:

$4a=4(3)=12$ .

Given, a parabola with equation

$x^2 =6y$

This is parabola of the form $x^2=4ay$ which opens upward.

So,

By comparing the given parabola equation with the standard equation, we get,

$4a=6$

$a=\frac{3}{2}$

Hence,

Coordinates of the focus :

$(0,a)=\left (0,\frac{3}{2}\right)$

Axis of the parabola:

It can be seen that the axis of this parabola is Y-Axis.

The equation of the directrix

$y=-a,\Rightarrow y=-\frac{3}{2}\Rightarrow y+\frac{3}{2}=0$

The length of the latus rectum:

$4a=4(\frac{3}{2})=6$ .

Given, a parabola with equation

$y^2 =-8x$

This is parabola of the form $y^2=-4ax$ which opens towards left.

So,

By comparing the given parabola equation with the standard equation, we get,

$-4a=-8$

$a=2$

Hence,

Coordinates of the focus :

$(-a,0)=(-2,0)$

Axis of the parabola:

It can be seen that the axis of this parabola is X-Axis.

The equation of the directrix

$x=a,\Rightarrow x=2\Rightarrow x-2=0$

The length of the latus rectum:

$4a=4(2)=8$ .

Given, a parabola with equation

$x^2 =-16y$

This is parabola of the form $x^2=-4ay$ which opens downwards.

So,

By comparing the given parabola equation with the standard equation, we get,

$-4a=-16$

$a=4$

Hence,

Coordinates of the focus :

$(0,-a)=(0,-4)$

Axis of the parabola:

It can be seen that the axis of this parabola is Y-Axis.

The equation of the directrix

$y=a,\Rightarrow y=4\Rightarrow y-4=0$

The length of the latus rectum:

$4a=4(4)=16$ .

Given, a parabola with equation

$y^2 =10x$

This is parabola of the form $y^2=4ax$ which opens towards the right.

So,

By comparing the given parabola equation with the standard equation, we get,

$4a=10$

$a=\frac{10}{4}=\frac{5}{2}$

Hence,

Coordinates of the focus :

$(a,0)=\left(\frac{5}{2},0\right)$

Axis of the parabola:

It can be seen that the axis of this parabola is X-Axis.

The equation of the directrix

$x=-a,\Rightarrow x=-\frac{5}{2}\Rightarrow x+\frac{5}{2}=0\Rightarrow 2x+5=0$

The length of the latus rectum:

$4a=4(\frac{5}{2})=10$ .

Given, a parabola with equation

$x^2 =-9y$

This is parabola of the form $x^2=-4ay$ which opens downwards.

So

By comparing the given parabola equation with the standard equation, we get,

$-4a=-9$

$a=\frac{9}{4}$

Hence,

Coordinates of the focus :

$(0,-a)=\left (0,-\frac{9}{4}\right)$

Axis of the parabola:

It can be seen that the axis of this parabola is Y-Axis.

The equation of the directrix

$y=a,\Rightarrow y=\frac{9}{4}\Rightarrow y-\frac{9}{4}=0$

The length of the latus rectum:

$4a=4\left(\frac{9}{4}\right)=9$ .

Given, in a parabola,

Focus : (6,0) And Directrix : $x = - 6$

Here,

Focus is of the form (a, 0), which means it lies on the X-axis. And Directrix is of the form $x=-a$ which means it lies left to the Y-Axis.

These are the condition when the standard equation of a parabola is. $y^2=4ax$

Hence the Equation of Parabola is

$y^2=4ax$

Here, it can be seen that:

$a=6$

Hence the Equation of the Parabola is:

$\Rightarrow y^2=4ax\Rightarrow y^2=4(6)x$

$\Rightarrow y^2=24x$ .

Given,in a parabola,

Focus : Focus (0,–3); directrix $y = 3$

Here,

Focus is of the form (0,-a), which means it lies on the Y-axis. And Directrix is of the form $y=a$ which means it lies above X-Axis.

These are the conditions when the standard equation of a parabola is $x^2=-4ay$ .

Hence the Equation of Parabola is

$x^2=-4ay$

Here, it can be seen that:

$a=3$

Hence the Equation of the Parabola is:

$\Rightarrow x^2=-4ay\Rightarrow x^2=-4(3)y$

$\Rightarrow x^2=-12y$ .

Given,

Vertex (0,0) And focus (3,0)

As vertex of the parabola is (0,0) and focus lies in the positive X-axis, The parabola will open towards the right, And the standard equation of such parabola is

$y^2=4ax$

Here it can be seen that $a=3$

So, the equation of a parabola is

$\Rightarrow y^2=4ax\Rightarrow y^2=4(3)x$

$\Rightarrow y^2=12x$ .

Given,

Vertex (0,0) And focus (-2,0)

As vertex of the parabola is (0,0) and focus lies in the negative X-axis, The parabola will open towards left, And the standard equation of such parabola is

$y^2=-4ax$

Here it can be seen that $a=2$

So, the equation of a parabola is

$\Rightarrow y^2=-4ax\Rightarrow y^2=-4(2)x$

$\Rightarrow y^2=-8x$ .

Given

The Vertex of the parabola is (0,0).

The parabola is passing through (2,3) and axis is along the x -axis, it will open towards right. and the standard equation of such parabola is

$y^2=4ax$

Now since it passes through (2,3)

$3^2=4a(2)$

$9=8a$

$a=\frac{8}{9}$

So the Equation of Parabola is ;

$\Rightarrow y^2=4\left(\frac{9}{8}\right)x$

$\Rightarrow y^2=\left(\frac{9}{2}\right)x$

$\Rightarrow 2y^2=9x$

Given a parabola,

with Vertex (0,0), passing through (5,2) and symmetric with respect to the y -axis.

Since the parabola is symmetric with respect to Y=axis, it's axis will ve Y-axis. and since it passes through the point (5,2), it must go through the first quadrant.

So the standard equation of such parabola is

$x^2=4ay$

Now since this parabola is passing through (5,2)

$5^2=4a(2)$

$25=8a$

$a=\frac{25}{8}$

Hence the equation of the parabola is

$\Rightarrow x^2=4\left ( \frac{25}{8} \right )y$

$\Rightarrow x^2=\left ( \frac{25}{2} \right )y$

$\Rightarrow 2x^2=25y$

NCERT class 11 maths chapter 11 question answer - Exercise: 11.3

Given

The equation of the ellipse

$\frac{x^2}{36} + \frac{y^2}{16} = 1$

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

We get

$a=6$ and $b=4$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{6^2-4^2}$

$c=\sqrt{20}=2\sqrt{5}$

Hence,

Coordinates of the foci:

$(c,0)\:and\:(-c,0)=(2\sqrt{5},0)\:and\:(-2\sqrt{5},0)$

The vertices:

$(a,0)\:and\:(-a,0)=(6,0)\:and\:(-6,0)$

The length of the major axis:

$2a=2(6)=12$

The length of minor axis:

$2b=2(4)=8$

The eccentricity :

$e=\frac{c}{a}=\frac{2\sqrt{5}}{6}=\frac{\sqrt{5}}{3}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(4)^2}{6}=\frac{32}{6}=\frac{16}{3}$

Given

The equation of the ellipse

$\frac{x^2}{4} + \frac{y^2}{25} =1$

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$

We get

$a=5$ and $b=2$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{5^2-2^2}$

$c=\sqrt{21}$

Hence,

Coordinates of the foci:

$(0,c)\:and\:(0,-c)=(0,\sqrt{21})\:and\:(0,-\sqrt{21})$

The vertices:

$(0,a)\:and\:(0,-a)=(0,5)\:and\:(0,-5)$

The length of the major axis:

$2a=2(5)=10$

The length of minor axis:

$2b=2(2)=4$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{21}}{6}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(2)^2}{5}=\frac{8}{5}$

Given

The equation of the ellipse

$\frac{x^2}{16} + \frac{y^2}{9} = 1$

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

We get

$a=4$ and $b=3$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{4^2-3^2}$

$c=\sqrt{7}$

Hence,

Coordinates of the foci:

$(c,0)\:and\:(-c,0)=(\sqrt{7},0)\:and\:(-\sqrt{7},0)$

The vertices:

$(a,0)\:and\:(-a,0)=(4,0)\:and\:(-4,0)$

The length of the major axis:

$2a=2(4)=8$

The length of minor axis:

$2b=2(3)=6$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{7}}{4}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(3)^2}{4}=\frac{18}{4}=\frac{9}{2}$

Given

The equation of the ellipse

$\frac{x^2}{25} + \frac{y^2}{100} = 1$

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$

We get

$a=10$ and $b=5$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{10^2-5^2}$

$c=\sqrt{75}=5\sqrt{3}$

Hence,

Coordinates of the foci:

$(0,c)\:and\:(0,-c)=(0,5\sqrt{3})\:and\:(0,-5\sqrt{3})$

The vertices:

$(0,a)\:and\:(0,-a)=(0,10)\:and\:(0,-10)$

The length of the major axis:

$2a=2(10)=20$

The length of minor axis:

$2b=2(5)=10$

The eccentricity :

$e=\frac{c}{a}=\frac{5\sqrt{3}}{10}=\frac{\sqrt{3}}{2}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(5)^2}{10}=\frac{50}{10}=5$

Given

The equation of ellipse

$\frac{x^2}{49} + \frac{y^2}{36} = 1\Rightarrow \frac{x^2}{7^2} + \frac{y^2}{6^2} = 1$

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with standard equation of ellipse, which is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

We get

$a=7$ and $b=6$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{7^2-6^2}$

$c=\sqrt{13}$

Hence,

Coordinates of the foci:

$(c,0)\:and\:(-c,0)=(\sqrt{13},0)\:and\:(-\sqrt{13},0)$

The vertices:

$(a,0)\:and\:(-a,0)=(7,0)\:and\:(-7,0)$

The length of major axis:

$2a=2(7)=14$

The length of minor axis:

$2b=2(6)=12$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{13}}{7}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(6)^2}{7}=\frac{72}{7}$

Given

The equation of the ellipse

$\frac{x^2}{100} + \frac{y^2}{400} =1\Rightarrow \frac{x^2}{10^2} + \frac{y^2}{20^2} =1$

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$

We get

$a=20$ and $b=10$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{20^2-10^2}$

$c=\sqrt{300}=10\sqrt{3}$

Hence,

Coordinates of the foci:

$(0,c)\:and\:(0,-c)=(0,10\sqrt{3})\:and\:(0,-10\sqrt{3})$

The vertices:

$(0,a)\:and\:(0,-a)=(0,20)\:and\:(0,-20)$

The length of the major axis:

$2a=2(20)=40$

The length of minor axis:

$2b=2(10)=20$

The eccentricity :

$e=\frac{c}{a}=\frac{10\sqrt{3}}{20}=\frac{\sqrt{3}}{2}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(10)^2}{20}=\frac{200}{20}=10$

Given

The equation of the ellipse

$36x^2 + 4y^2 =144$

$\Rightarrow \frac{36}{144}x^2 + \frac{4}{144}y^2 =1$

$\Rightarrow \frac{1}{4}x^2 + \frac{1}{36}y^2 =1$

$\frac{x^2}{2^2} + \frac{y^2}{6^2} = 1$

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$

We get

$a=6$ and $b=2$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{6^2-2^2}$

$c=\sqrt{32}=4\sqrt{2}$

Hence,

Coordinates of the foci:

$(0,c)\:and\:(0,-c)=(0,4\sqrt{2})\:and\:(0,-4\sqrt{2})$

The vertices:

$(0,a)\:and\:(0,-a)=(0,6)\:and\:(0,-6)$

The length of the major axis:

$2a=2(6)=12$

The length of minor axis:

$2b=2(2)=4$

The eccentricity :

$e=\frac{c}{a}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(2)^2}{6}=\frac{8}{6}=\frac{4}{3}$

Given

The equation of the ellipse

$16x^2 + y^2 = 16$

$\frac{16x^2}{16} + \frac{y^2}{16} = 1$

$\frac{x^2}{1^2} + \frac{y^2}{4^2} = 1$

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$

We get

$a=4$ and $b=1$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{4^2-1^2}$

$c=\sqrt{15}$

Hence,

Coordinates of the foci:

$(0,c)\:and\:(0,-c)=(0,\sqrt{15})\:and\:(0,-\sqrt{15})$

The vertices:

$(0,a)\:and\:(0,-a)=(0,4)\:and\:(0,-4)$

The length of the major axis:

$2a=2(4)=8$

The length of minor axis:

$2b=2(1)=2$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{15}}{4}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(1)^2}{4}=\frac{2}{4}=\frac{1}{2}$

Given

The equation of the ellipse

$4x^2 + 9y^2 =36$

$\frac{4x^2}{36} + \frac{9y^2}{36} = 1$

$\frac{x^2}{9} + \frac{y^2}{4} = 1$

$\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

We get

$a=3$ and $b=2$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{3^2-2^2}$

$c=\sqrt{5}$

Hence,

Coordinates of the foci:

$(c,0)\:and\:(-c,0)=(\sqrt{5},0)\:and\:(-\sqrt{5},0)$

The vertices:

$(a,0)\:and\:(-a,0)=(3,0)\:and\:(-3,0)$

The length of the major axis:

$2a=2(3)=6$

The length of minor axis:

$2b=2(2)=4$

The eccentricity :

$e=\frac{c}{a}=\frac{\sqrt{5}}{3}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(2)^2}{3}=\frac{8}{3}$

Given, In an ellipse,

Vertices (± 5, 0), foci (± 4, 0)

Here Vertices and focus of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( vertices and foci) with the given one, we get

$a=5$ and $c=4$

Now, As we know the relation,

$a^2=b^2+c^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}$

$b=\sqrt{5^2-4^2}$

$b=\sqrt{9}$

$b=3$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

Which is

$\frac{x^2}{25}+\frac{y^2}{9}=1$ .

Given, In an ellipse,

Vertices (0, ± 13), foci (0, ± 5)

Here Vertices and focus of the ellipse are in Y-axis so the major axis of this ellipse will be Y-axis.

Therefore, the equation of the ellipse will be of the form:

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( vertices and foci) with the given one, we get

$a=13$ and $c=5$

Now, As we know the relation,

$a^2=b^2+c^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}$

$b=\sqrt{13^2-5^2}$

$b=\sqrt{169-25}$

$b=\sqrt{144}$

$b=12$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{12^2}+\frac{y^2}{13^3}=1$

Which is

$\frac{x^2}{144}+\frac{y^2}{169}=1$ .

Given, In an ellipse,

Vertices (± 6, 0), foci (± 4, 0)

Here Vertices and focus of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( vertices and foci) with the given one, we get

$a=6$ and $c=4$

Now, As we know the relation,

$a^2=b^2+c^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}$

$b=\sqrt{6^2-4^2}$

$b=\sqrt{36-16}$

$b=\sqrt{20}$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{6^2}+\frac{y^2}{(\sqrt{20})^2}=1$

Which is

$\frac{x^2}{36}+\frac{y^2}{20}=1$ .

Given, In an ellipse,

Ends of the major axis (± 3, 0), ends of minor axis (0, ± 2)

Here, the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( ends of the major and minor axis ) with the given one, we get

$a=3$ and $b=2$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{3^2}+\frac{y^2}{2^2}=1$

Which is

$\frac{x^2}{9}+\frac{y^2}{4}=1$ .

Given, In an ellipse,

Ends of the major axis (0, ± $\sqrt{5}$ ), ends of minor axis (± 1, 0)

Here, the major axis of this ellipse will be Y-axis.

Therefore, the equation of the ellipse will be of the form:

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( ends of the major and minor axis ) with the given one, we get

$a=\sqrt{5}$ and $b=1$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{1^2}+\frac{y^2}{(\sqrt{5})^2}=1$

Which is

$\frac{x^2}{1}+\frac{y^2}{5}=1$ .

Given, In an ellipse,

Length of major axis 26, foci (± 5, 0)

Here, the focus of the ellipse is in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( Length of semimajor axis and foci) with the given one, we get

$2a=26\Rightarrow a=13$ and $c=5$

Now, As we know the relation,

$a^2=b^2+c^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}$

$b=\sqrt{13^2-5^2}$

$b=\sqrt{144}$

$b=12$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{13^2}+\frac{y^2}{12^2}=1$

Which is

$\frac{x^2}{169}+\frac{y^2}{144}=1$ .

Given, In an ellipse,

Length of minor axis 16, foci (0, ± 6).

Here, the focus of the ellipse is on the Y-axis so the major axis of this ellipse will be Y-axis.

Therefore, the equation of the ellipse will be of the form:

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( length of semi-minor axis and foci) with the given one, we get

$2b=16\Rightarrow b=8$ and $c=6$

Now, As we know the relation,

$a^2=b^2+c^2$

$a=\sqrt{b^2+c^2}$

$a=\sqrt{8^2+6^2}$

$a=\sqrt{64+36}$

$a=\sqrt{100}$

$a=10$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{8^2}+\frac{y^2}{10^3}=1$

Which is

$\frac{x^2}{64}+\frac{y^2}{100}=1$ .

Given, In an ellipse,

V Foci (± 3, 0), a = 4

Here foci of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( vertices and foci) with the given one, we get

$a=4$ and $c=3$

Now, As we know the relation,

$a^2=b^2+c^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}$

$b=\sqrt{4^2-3^2}$

$b=\sqrt{7}$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{4^2}+\frac{y^2}{(\sqrt{7})^2}=1$

Which is

$\frac{x^2}{16}+\frac{y^2}{7}=1$ .

Given,In an ellipse,

b = 3, c = 4, centre at the origin; foci on the x axis.

Here foci of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

Also Given,

$b=3$ and $c=4$

Now, As we know the relation,

$a^2=b^2+c^2$

$a^2=3^2+4^2$

$a^2=25$

$a=5$

Hence, The Equation of the ellipse will be :

$\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

Which is

$\frac{x^2}{25}+\frac{y^2}{9}=1$ .

Given,in an ellipse

Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).

Since, The major axis of this ellipse is on the Y-axis, the equation of the ellipse will be of the form:

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

Now since the ellipse passes through points,(3, 2)

$\frac{3^2}{b^2}+\frac{2^2}{a^2}=1$

${9a^2+4b^2}={a^2b^2}$

since the ellipse also passes through points,(1, 6).

$\frac{1^2}{b^2}+\frac{6^2}{a^2}=1$

$a^2+36b^2=a^2b^2$

On solving these two equation we get

$a^2=40$ and $b^2=10$

Thus, The equation of the ellipse will be

$\frac{x^2}{10}+\frac{y^2}{40}=1$ .

Given, in an ellipse

Major axis on the x-axis and passes through the points (4,3) and (6,2).

Since The major axis of this ellipse is on the X-axis, the equation of the ellipse will be of the form:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Where $a$ and $b$ are the length of the semimajor axis and semiminor axis respectively.

Now since the ellipse passes through the point,(4,3)

$\frac{4^2}{a^2}+\frac{3^2}{b^2}=1$

${16b^2+9a^2}={a^2b^2}$

since the ellipse also passes through the point (6,2).

$\frac{6^2}{a^2}+\frac{2^2}{b^2}=1$

$4a^2+36b^2=a^2b^2$

On solving this two equation we get

$a^2=52$ and $b^2=13$

Thus, The equation of the ellipse will be

$\frac{x^2}{52}+\frac{y^2}{13}=1$

NCERT class 11 maths chapter 11 question answer - Exercise: 11.4

Given a Hyperbola equation,

$\frac{x^2}{16} - \frac{y^2}{9} = 1$

Can also be written as

$\frac{x^2}{4^2} - \frac{y^2}{3^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

We get,

$a=4$ and $b=3$

Now, As we know the relation in a hyperbola,

$c^2=a^2+b^2$

$c=\sqrt{a^2+b^2}$

$c=\sqrt{4^2+3^2}$

$c=5$

Here as we can see from the equation that the axis of the hyperbola is X -axis. So,

Coordinates of the foci:

$(c,0) \:and\:(-c,0)=(5,0)\:and\:(-5,0)$

The Coordinates of vertices:

$(a,0) \:and\:(-a,0)=(4,0)\:and\:(-4,0)$

The Eccentricity:

$e=\frac{c}{a}=\frac{5}{4}$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(3)^2}{4}=\frac{18}{4}=\frac{9}{2}$

Given a Hyperbola equation,

$\frac{y^2}{9} - \frac{x^2}{27} = 1$

Can also be written as

$\frac{y^2}{3^2} - \frac{x^2}{(\sqrt{27})^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

We get,

$a=3$ and $b=\sqrt{27}$

Now, As we know the relation in a hyperbola,

$c^2=a^2+b^2$

$c=\sqrt{a^2+b^2}$

$c=\sqrt{3^2+(\sqrt{27})^2}$

$c=\sqrt{36}$

$c=6$

Here as we can see from the equation that the axis of the hyperbola is Y-axis. So,

Coordinates of the foci:

$(0,c) \:and\:(0,-c)=(0,6)\:and\:(0,-6)$

The Coordinates of vertices:

$(0,a) \:and\:(0,-a)=(0,3)\:and\:(0,-3)$

The Eccentricity:

$e=\frac{c}{a}=\frac{6}{3}=2$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(27)}{3}=\frac{54}{3}=18$

Given a Hyperbola equation,

$9 y^2 - 4 x^2 =36$

Can also be written as

$\frac{9y^2}{36} - \frac{4x^2}{36} = 1$

$\frac{y^2}{2^2} - \frac{x^2}{3^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

We get,

$a=2$ and $b=3$

Now, As we know the relation in a hyperbola,

$c^2=a^2+b^2$

$c=\sqrt{a^2+b^2}$

$c=\sqrt{2^2+3^2}$

$c=\sqrt{13}$

Hence,

Coordinates of the foci:

$(0,c) \:and\:(0,-c)=(0,\sqrt{13})\:and\:(0,-\sqrt{13})$

The Coordinates of vertices:

$(0,a) \:and\:(0,-a)=(0,2)\:and\:(0,-2)$

The Eccentricity:

$e=\frac{c}{a}=\frac{\sqrt{13}}{2}$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(9)}{2}=\frac{18}{2}=9$

Given a Hyperbola equation,

$16x^2 - 9y^2 = 576$

Can also be written as

$\frac{16x^2}{576} - \frac{9y^2}{576} = 1$

$\frac{x^2}{36} - \frac{y^2}{64} = 1$

$\frac{x^2}{6^2} - \frac{y^2}{8^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

We get,

$a=6$ and $b=8$

Now, As we know the relation in a hyperbola,

$c^2=a^2+b^2$

$c=\sqrt{a^2+b^2}$

$c=\sqrt{6^2+8^2}$

$c=10$

Therefore,

Coordinates of the foci:

$(c,0) \:and\:(-c,0)=(10,0)\:and\:(-10,0)$

The Coordinates of vertices:

$(a,0) \:and\:(-a,0)=(6,0)\:and\:(-6,0)$

The Eccentricity:

$e=\frac{c}{a}=\frac{10}{6}=\frac{5}{3}$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(8)^2}{6}=\frac{128}{6}=\frac{64}{3}$

Given a Hyperbola equation,

$5y^2 - 9x^2 = 36$

Can also be written as

$\frac{5y^2}{36} - \frac{9x^2}{36} = 1$

$\frac{y^2}{\frac{36}{5}} - \frac{x^2}{4} = 1$

$\frac{y^2}{(\frac{6}{\sqrt{5}})^2} - \frac{x^2}{2^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

We get,

$a=\frac{6}{\sqrt{5}}$

and $b=2$

Now, As we know the relation in a hyperbola,

$c^2=a^2+b^2$

$c=\sqrt{a^2+b^2}$

$c=\sqrt{(\frac{6}{\sqrt{5}})^2+2^2}$

$c=\sqrt{\frac{56}{5}}$

$c=2\sqrt{\frac{14}{5}}$

Here as we can see from the equation that the axis of the hyperbola is Y-axis. So,

Coordinates of the foci:

$(0,c) \:and\:(0,-c)=\left(0,2\sqrt{\frac{14}{5}}\right)\:and\:\left(0,-2\sqrt{\frac{14}{5}}\right)$

The Coordinates of vertices:

$(0,a) \:and\:(0,-a)=\left(0,\frac{6}{\sqrt{5}}\right)\:and\:\left(0,-\frac{6}{\sqrt{5}}\right)$

The Eccentricity:

$e=\frac{c}{a}=\frac{2\sqrt{\frac{14}{5}}}{\frac{6}{\sqrt{5}}}=\frac{\sqrt{14}}{3}$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(4)}{\frac{6}{\sqrt{5}}}=\frac{4\sqrt{5}}{3}$

Given a Hyperbola equation,

$49y^2 - 16x^2 = 784$

Can also be written as

$\frac{49y^2}{784} - \frac{16x^2}{784} = 1$

$\frac{y^2}{16} - \frac{x^2}{49} = 1$

$\frac{y^2}{4^2} - \frac{x^2}{7^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

We get,

$a=4$ and $b=7$

Now, As we know the relation in a hyperbola,

$c^2=a^2+b^2$

$c=\sqrt{a^2+b^2}$

$c=\sqrt{4^2+7^2}$

$c=\sqrt{65}$

Therefore,

Coordinates of the foci:

$(0,c) \:and\:(0,-c)=(0,\sqrt{65})\:and\:(0,-\sqrt{65})$

The Coordinates of vertices:

$(0,a) \:and\:(0,-a)=(0,4)\:and\:(0,-4)$

The Eccentricity:

$e=\frac{c}{a}=\frac{\sqrt{65}}{4}$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(49)}{4}=\frac{98}{4}=\frac{49}{2}$

Given, in a hyperbola

Vertices (± 2, 0), foci (± 3, 0)

Here, Vertices and focii are on the X-axis so, the standard equation of the Hyperbola will be ;

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

By comparing the standard parameter (Vertices and foci) with the given one, we get

$a=2$ and $c=3$

Now, As we know the relation in a hyperbola

$c^2=a^2+b^2$

$b^2=c^2-a^2$

$b^2=3^2-2^2$

$b^2=9-4=5$

Hence,The Equation of the hyperbola is ;

$\frac{x^2}{4}-\frac{y^2}{5}=1$

Given, in a hyperbola

Vertices (0, ± 5), foci (0, ± 8)

Here, Vertices and focii are on the Y-axis so, the standard equation of the Hyperbola will be ;

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

By comparing the standard parameter (Vertices and foci) with the given one, we get

$a=5$ and $c=8$

Now, As we know the relation in a hyperbola

$c^2=a^2+b^2$

$b^2=c^2-a^2$

$b^2=8^2-5^2$

$b^2=64-25=39$

Hence, The Equation of the hyperbola is ;

$\frac{y^2}{25}-\frac{x^2}{39}=1$ .

Given, in a hyperbola

Vertices (0, ± 3), foci (0, ± 5)

Here, Vertices and focii are on the Y-axis so, the standard equation of the Hyperbola will be ;

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

By comparing the standard parameter (Vertices and foci) with the given one, we get

$a=3$ and $c=5$

Now, As we know the relation in a hyperbola

$c^2=a^2+b^2$

$b^2=c^2-a^2$

$b^2=5^2-3^2$

$b^2=25-9=16$

Hence, The Equation of the hyperbola is ;

$\frac{y^2}{9}-\frac{x^2}{16}=1$ .

Given, in a hyperbola

Foci (± 5, 0), the transverse axis is of length 8.

Here, focii are on the X-axis so, the standard equation of the Hyperbola will be ;

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

By comparing the standard parameter (transverse axis length and foci) with the given one, we get

$2a=8\Rightarrow a=4$ and $c=5$

Now, As we know the relation in a hyperbola

$c^2=a^2+b^2$

$b^2=c^2-a^2$

$b^2=5^2-4^2$

$b^2=25-16=9$

Hence, The Equation of the hyperbola is ;

$\frac{x^2}{16}-\frac{y^2}{9}=1$

Given, in a hyperbola

Foci (0, ±13), the conjugate axis is of length 24.

Here, focii are on the Y-axis so, the standard equation of the Hyperbola will be ;

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

By comparing the standard parameter (length of conjugate axis and foci) with the given one, we get

$2b=24\Rightarrow b=12$ and $c=13$

Now, As we know the relation in a hyperbola

$c^2=a^2+b^2$

$a^2=c^2-b^2$

$a^2=13^2-12^2$

$a^2=169-144=25$

Hence, The Equation of the hyperbola is ;

$\frac{y^2}{25}-\frac{x^2}{144}=1$ .

Given, in a hyperbola

Foci $(\pm 3\sqrt5, 0)$ , the latus rectum is of length 8.

Here, focii are on the X-axis so, the standard equation of the Hyperbola will be ;

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

By comparing standard parameter (length of latus rectum and foci) with the given one, we get

$c=3\sqrt{5}$ and

$\frac{2b^2}{a}=8\Rightarrow 2b^2=8a\Rightarrow b^2=4a$

Now, As we know the relation in a hyperbola

$c^2=a^2+b^2$

$c^2=a^2+4a$

$a^2+4a=(3\sqrt{5})^2$

$a^2+4a=45$

$a^2+9a-5a-45=0$

$(a+9)(a-5)=0$

$a=-9\:or\:5$

Since $a$ can never be negative,

$a=5$

$a^2=25$

$b^2=4a=4(5)=20$

Hence, The Equation of the hyperbola is ;

$\frac{x^2}{25}-\frac{y^2}{20}=1$

Given, in a hyperbola

Foci (± 4, 0), the latus rectum is of length 12

Here, focii are on the X-axis so, the standard equation of the Hyperbola will be ;

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

By comparing standard parameter (length of latus rectum and foci) with the given one, we get

$c=4$ and

$\frac{2b^2}{a}=12\Rightarrow 2b^2=12a\Rightarrow b^2=6a$

Now, As we know the relation in a hyperbola

$c^2=a^2+b^2$

$c^2=a^2+6a$

$a^2+6a=4^2$

$a^2+6a=16$

$a^2+8a-2a-16=0$

$(a+8)(a-2)=0$

$a=-8\:or\:2$

Since $a$ can never be negative,

$a=2$

$a^2=4$

$b^2=6a=6(2)=12$

Hence, The Equation of the hyperbola is ;

$\frac{x^2}{4}-\frac{y^2}{12}=1$

Given, in a hyperbola

vertices (± 7,0), And

$e = \frac{4}{3}$

Here, Vertices is on the X-axis so, the standard equation of the Hyperbola will be ;

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

By comparing the standard parameter (Vertices and eccentricity) with the given one, we get

$a=7$ and

$e=\frac{c}{a}=\frac{c}{7}=\frac{4}{3}$

From here,

$c=\frac{28}{3}$

Now, As we know the relation in a hyperbola

$c^2=a^2+b^2$

$b^2=c^2-a^2$

$b^2=\left(\frac{28}{3}\right)^2-7^2$

$b^2=\left(\frac{784}{9}\right)-49$

$b^2=\left(\frac{784-441}{9}\right)=\frac{343}{9}$

Hence, The Equation of the hyperbola is ;

$\frac{x^2}{49}-\frac{9y^2}{343}=1$

Given, in a hyperbola,

Foci $(0,\pm\sqrt{10})$ , passing through (2,3)

Since foci of the hyperbola are in Y-axis, the equation of the hyperbola will be of the form ;

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

By comparing standard parameter (foci) with the given one, we get

$c=\sqrt{10}$

Now As we know, in a hyperbola

$a^2+b^2=c^2$

$a^2+b^2=10\:\:\:\:\:\:\:....(1)$

Now As the hyperbola passes through the point (2,3)

$\frac{3^2}{a^2}-\frac{2^2}{b^2}=1$

$9b^2-4a^2=a^2b^2\:\;\;\:\:\;\:....(2)$

Solving Equation (1) and (2)

$9(10-a^2)-4a^2=a^2(10-a^2)$

$a^4-23a^2+90=0$

$(a^2)^2-18a^2-5a^2+90=0$

$(a^2-18)(a^2-5)=0$

$a^2=18\:or\:5$

Now, as we know that in a hyperbola $c$ is always greater than, $a$ we choose the value

$a^2=5$

$b^2=10-a^2=10-5=5$

Hence The Equation of the hyperbola is

$\frac{y^2}{5}-\frac{x^2}{5}=1$

Class 11 maths chapter 11 NCERT solutions - Miscellaneous Exercise

Le the parabolic reflector opens towards the right.

So the equation of parabolic reflector will be,

$y^2=4ax$

Now, Since this curve will pass through the point (5,10) if we assume origin at the optical centre,

So

$10^2=4a(5)$

$a=\frac{100}{20}=5$

Hence, The focus of the parabola is,

$(a,0)=(5,0)$ .

Alternative Method,

As we know on any concave curve

$f=\frac{R}{2}$

Hence, Focus

$f=\frac{R}{2}=\frac{10}{2}=5$ .

Hence the focus is 5 cm right to the optical centre.

Since the Axis of the parabola is vertical, Let the equation of the parabola be,

$x^2=4ay$

it can be seen that this curve will pass through the point (5/2, 10) if we assume origin at the bottom end of the parabolic arch.

So,

$\left(\frac{5}{2}\right)^2=4a(10)$

$a=\frac{25}{160}=\frac{5}{32}$

Hence, the equation of the parabola is

$x^2=4\times\frac{5}{32}\times y$

$x^2=\frac{20}{32} y$

$x^2=\frac{5}{8} y$

Now, when y = 2 the value of x will be

$x=\sqrt{(\frac{5}{8}\times2)}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$

Hence the width of the arch at this height is

$2x=2\times\frac{\sqrt{5}}{2}=\sqrt{5}.$

Given,

The width of the parabolic cable = 100m

The length of the shorter supportive wire attached = 6m

The length of the longer supportive wire attached = 30m

Since the rope opens towards upwards, the equation will be of the form

$x^2=4ay$

Now if we consider origin at the centre of the rope, the equation of the curve will pass through points, (50,30-6)=(50,24)

$24^2=4a50$

$a=\frac{625}{24}$

Hence the equation of the parabola is

$x^2=4\times \frac{625}{24}\times y$

$x^2= \frac{625}{6}\times y$

Now at a point, 18 m right from the centre of the rope, the x coordinate of that point will be 18, so by the equation, the y-coordinate will be

$y=\frac{x^2}{4a}=\frac{18^2}{4\times \frac{625}{6}}\approx 3.11m$

Hence the length of the supporting wire attached to roadway from the middle is 3.11+6=9.11m.

The equation of the semi-ellipse will be of the form

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\:,y>0$

Now, According to the question,

the length of major axis = 2a = 8 $\Rightarrow a=4$

The length of the semimajor axis =2 $\Rightarrow b=2$

Hence the equation will be,

$\frac{x^2}{4^2}+\frac{y^2}{2^2}=1\:,y>0$

$\frac{x^2}{16}+\frac{y^2}{4}=1\:,y>0$

Now, at point 1.5 cm from the end, the x coordinate is 4-1.5 = 2.5

So, the height at this point is

$\frac{(2.5)^2}{16}+\frac{y^2}{4}=1\Rightarrow y=\sqrt{4(1-\frac{2.5^2}{16})}$

$y\approx 1.56m$

Hence the height of the required point is 1.56 m.

Let $\theta$ be the angle that rod makes with the ground,

Now, at a point 3 cm from the end,

$\cos\theta=\frac{x}{9}$

At the point touching the ground

$\sin\theta=\frac{y}{3}$

Now, As we know the trigonometric identity,

$\sin^2\theta+\cos^2\theta=1$

$\left (\frac{x}{9} \right )^2+\left ( \frac{y}{3} \right )^2=1$

$\frac{x^2}{81}+\frac{y^2}{9}=1$

Hence the equation is,

$\frac{x^2}{81}+\frac{y^2}{9}=1$

Given the parabola,

$x^2=12y$

Comparing this equation with $x^2=4ay$ , we get

$a=3$

Now, As we know the coordinates of ends of latus rectum are:

$(2a,a)\:and\:(-2a,a)$

So, the coordinates of latus rectum are,

$(2a,a)\:and\:(-2a,a)=(6,3)\:and\:(-6,3)$

Now the area of the triangle with coordinates (0,0),(6,3) and (-6,3)

Widht of the triangle = 2*6=12

Height of the triangle = 3

So The area =

$\frac{1}{2}\times base\times height=\frac{1}{2}\times12\times3=18$

Hence the required area is 18 unit square.

As we know that if a point moves in a plane in such a way that its distance from two-point remain constant then the path is an ellipse.

Now, According to the question,

the distance between the point from where the sum of the distance from a point is constant = 10

$\Rightarrow 2a=10\Rightarrow a=5$

Now, the distance between the foci=8

$\Rightarrow 2c=8\Rightarrow c=4$

Now, As we know the relation,

$c^2=a^2-b^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}=\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3$

Hence the equation of the ellipse is,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

$\Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1$

Hence the path of the man will be

$\Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1$

Given, an equilateral triangle inscribed in parabola with the equation. $y^2 = 4 ax$

The one coordinate of the triangle is A(0,0).

Now, let the other two coordinates of the triangle are

$B(x,\sqrt{4ax})$ and $C(x,-\sqrt{4ax})$

Now, Since the triangle is equilateral,

$BC=AB=CA$

$2\sqrt{4ax}=\sqrt{(x-0)^2+(\sqrt{4ax}-0)^2}$

$x^2=12ax$

$x=12a$

The coordinates of the points of the equilateral triangle are,

$(0,0),(12,\sqrt{4a\times 12a}),(12,-\sqrt{4a\times 12a})=(0,0),(12,4\sqrt{3}a)\:and\:(12,-4\sqrt{3}a)$

So, the side of the triangle is

$2\sqrt{4ax}=2\times4\sqrt{3}a=8\sqrt{3}a$

## 11.1 Introduction

11.2 Sections of a Cone

11.3 Circle

11.4 Parabola

11.5 Ellipse

11.6 Hyperbola

• Circle- It is the set of all points in the plane that are equidistant(or the same distance) from a fixed point in the plane. The fixed point is the centre of the circle and the distance from the centre to a point on the circle is the radius of the circle. The equation of a circle with centre (h, k) and the radius r is - $\dpi{150} \(x-h)^2+(y-k)^2=r^2$
• Parabola- It is the set of all points in the plane that are equidistant(or the same distance) from a fixed line and a fixed point (not on the line) in the plane. The fixed point F is the focus of the parabola and the fixed line is called the directrix of the parabola. If the coordinates of focus(F) is (a, 0) a > 0 and directrix x = – a, then the equation of the parabola is - $\dpi{150} y^2=4ax$
• Ellipse- It is formed by a point, which moves in a plane in such a manner that the sum of its distances from two fixed points in a plane is constant. The two fixed points in the plane are called the foci of the ellipse. If the foci of the ellipse are on the x-axis, then the equation of an ellipse is - $\dpi{150} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
• Hyperbola- It is the locus of a point in a plane, which moves in such a manner so that the difference of whose distances from two fixed points in the plane is constant. The two fixed points in the hyperbola are called the foci of the hyperbola. If the foci of the hyperbola are on the x-axis, then the equation of a hyperbola is - $\dpi{150} \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Interested students can practice class 11 maths ch 11 question answer using the exercises listed below.

## NCERT Solutions For Class 11 Mathematics

 chapter-1 Sets chapter-2 Relations and Functions chapter-3 Trigonometric Functions chapter-4 Principle of Mathematical Induction chapter-5 Complex Numbers and Quadratic equations chapter-6 Linear Inequalities chapter-7 Permutation and Combinations chapter-8 Binomial Theorem chapter-9 Sequences and Series chapter-10 Straight Lines chapter-11 Conic Section chapter-12 Introduction to Three Dimensional Geometry chapter-13 Limits and Derivatives chapter-14 Mathematical Reasoning chapter-15 Statistics chapter-16

## Key Features Of class 11 maths chapter 11 NCERT solutions

Easy Solutions: NCERT Solutions of ch 11 maths class 11 provide step-by-step solutions to all the questions in the textbook. These solutions are easy to understand and follow, making it easier for students to learn.

Simple Language: The language used in the textbook and chapter 11 class 11 maths solutions is simple and easy to comprehend. This helps students to grasp the concepts better and avoid confusion.

Extensive Coverage: The class 11 chapter 11 maths covers all the essential topics related to Conic Sections, such as the properties and equations of Circle, Parabola, Hyperbola, and Ellipse. This comprehensive coverage enables students to gain a deep understanding of the subject matter.

## NCERT Solutions For Class 11 - Subject Wise

Tip- You should remember standard equations and formulas for all the standard curves and try to solve problems from NCERT including miscellaneous exercise. If you are not able to do, you can take help from the NCERT solutions for class 11 maths chapter 11 conic section.

## NCERT Books and NCERT Syllabus

1. What is the number of topics covered in conic sections class 11 NCERT solutions?

Conic sections class 11 solutions encompass significant topics that enhance understanding of concepts such as ellipse, hyperbola, and more. The introductory portion of this chapter offers students an overall understanding of the basics that are crucial for board exams. Subsequent sections elaborate on elements like standard equations of parabola, degenerate conic sections, latus rectum, and ellipse.

2. How to score good marks in class 11 maths ncert solutions chapter 11?

Class 11 maths conic sections holds significant importance in Class 11 Mathematics. After students have covered the fundamental areas, they can tackle the exercise problems with greater ease. By solving problems pertaining to this chapter, students can sharpen their analytical and logical thinking abilities. Selecting appropriate study materials is also crucial for scoring well in Class 11 exams, including the board exams. The primary objective of developing NCERT Solutions is to assist students in identifying the concepts in which they need improvement and guiding them to achieve better scores.

3. Are the solutions provided by Careers360 for class 11 maths chapter 11 solutions in accordance with the CBSE syllabus?

Accuracy is crucial in Mathematics, and consistent practice is essential to achieve it. With numerous study materials available online, selecting the right one can be a daunting task. By choosing solutions that concentrate solely on the CBSE Syllabus, students can comprehend the crucial concepts that are significant from an examination perspective. Class 11th conic section solution provided by Careers360 team are includes all these points.

4. Where can I find the complete class 11 conic section solution ?

Students can get the detailed NCERT solutions for class 11 maths  by clicking on the link. students can practice these solutions to get indepth understanding of concepts.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

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 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

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 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

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 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

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 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

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 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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##### Architect

Individuals in the architecture career are the building designers who plan the whole construction keeping the safety and requirements of the people. Individuals in architect career in India provides professional services for new constructions, alterations, renovations and several other activities. Individuals in architectural careers in India visit site locations to visualize their projects and prepare scaled drawings to submit to a client or employer as a design. Individuals in architecture careers also estimate build costs, materials needed, and the projected time frame to complete a build.

2 Jobs Available
##### Landscape Architect

Having a landscape architecture career, you are involved in site analysis, site inventory, land planning, planting design, grading, stormwater management, suitable design, and construction specification. Frederick Law Olmsted, the designer of Central Park in New York introduced the title “landscape architect”. The Australian Institute of Landscape Architects (AILA) proclaims that "Landscape Architects research, plan, design and advise on the stewardship, conservation and sustainability of development of the environment and spaces, both within and beyond the built environment". Therefore, individuals who opt for a career as a landscape architect are those who are educated and experienced in landscape architecture. Students need to pursue various landscape architecture degrees, such as M.Des, M.Plan to become landscape architects. If you have more questions regarding a career as a landscape architect or how to become a landscape architect then you can read the article to get your doubts cleared.

2 Jobs Available
##### Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
##### Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 Jobs Available
##### Carpenter

Carpenters are typically construction workers. They stay involved in performing many types of construction activities. It includes cutting, fitting and assembling wood.  Carpenters may help in building constructions, bridges, big ships and boats. Here, in the article, we will discuss carpenter career path, carpenter salary, how to become a carpenter, carpenter job outlook.

2 Jobs Available
##### Welder

An individual who opts for a career as a welder is a professional tradesman who is skilled in creating a fusion between two metal pieces to join it together with the use of a manual or fully automatic welding machine in their welder career path. It is joined by intense heat and gas released between the metal pieces through the welding machine to permanently fix it.

2 Jobs Available
##### Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems.

2 Jobs Available
##### Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
##### Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 Jobs Available
##### Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
##### Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available
##### Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

3 Jobs Available
##### Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
##### Dentist

Those who wish to make a dentist career in India must know that dental training opens up a universe of expert chances. Notwithstanding private practice, the present dental school graduates can pick other dental profession alternatives, remembering working in medical clinic crisis rooms, leading propelled lab examinations, teaching future dental specialists, or in any event, venturing to the far corners of the planet with International health and relief organizations.

2 Jobs Available
##### Health Inspector

Individuals following a career as health inspectors have to face resistance and lack of cooperation while working on the sites. The health inspector's job description includes taking precautionary measures while inspecting to save themself from any external injury and the need to cover their mouth to avoid toxic substances. A health inspector does the desk job as well as the fieldwork. Health inspector jobs require one to travel long hours to inspect a particular place.

2 Jobs Available
##### Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
##### Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
##### Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
##### Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
##### Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story.

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
##### Fashion Blogger

Fashion bloggers use multiple social media platforms to recommend or share ideas related to fashion. A fashion blogger is a person who writes about fashion, publishes pictures of outfits, jewellery, accessories. Fashion blogger works as a model, journalist, and a stylist in the fashion industry. In current fashion times, these bloggers have crossed into becoming a star in fashion magazines, commercials, or campaigns.

2 Jobs Available
##### Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
##### Fashion Journalist

Fashion journalism involves performing research and writing about the most recent fashion trends. Journalists obtain this knowledge by collaborating with stylists, conducting interviews with fashion designers, and attending fashion shows, photoshoots, and conferences. A fashion Journalist  job is to write copy for trade and advertisement journals, fashion magazines, newspapers, and online fashion forums about style and fashion.

2 Jobs Available
##### Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available
##### QA Manager

Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes.

2 Jobs Available

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

2 Jobs Available
##### Reliability Engineer

Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment.

2 Jobs Available
##### Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### IT Consultant

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

2 Jobs Available
##### Data Architect

A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

2 Jobs Available
##### Security Engineer

The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

2 Jobs Available
##### UX Architect

A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

2 Jobs Available