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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

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 NCERT solutions 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.

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This Story also Contains
  1. Conic Sections Class 11 Questions And Answers
  2. Conic Sections Class 11 Questions And Answers PDF Free Download
  3. Conic Sections Class 11 Solutions - Important Formulae
  4. Conic Sections Class 11 NCERT Solutions (Intext Questions and Exercise)
  5. Maths chapter 11 class 11 Conic Sections - Topics
  6. 11.1 Introduction
  7. NCERT Solutions For Class 11 Mathematics
  8. Key Features Of class 11 maths chapter 11 NCERT solutions
  9. NCERT Solutions For Class 11 - Subject Wise
  10. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

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.

Also Read| Conic Section Class 11 Maths Chapter Notes

Also Read| NCERT Exemplar Solutions For Class 11 Maths Chapter Conic Sections

Conic Sections Class 11 Questions And Answers PDF Free Download

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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

centre (0,2) and radius 2

Answer:

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

centre (–2,3) and radius 4

Answer:

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

centre \left(\frac{1}{2},\frac{1}{4} \right ) and radius \frac{1}{12}

Answer:

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

centre (1,1) and radius \sqrt2

Answer:

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

centre (-a,-b) and radius \sqrt{a^2 - b^2}

Answer:

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

Question:6 Find the centre and radius of the circles.

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

Answer:

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.

Question:7 Find the centre and radius of the circles.

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

Answer:

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} .

Question:8 Find the centre and radius of the circles.

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

Answer:

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} .

Question:9 Find the centre and radius of the circles.

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

Answer:

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} .

Question:10 Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16 .

Answer:

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

Question:11 Find the equation of the circle passing through the points (2,3) and (–1,1) and hose centre is on the line x - 3y - 11 = 0 .

Answer:

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

Question:12 Find the equation of the circle with radius 5 whose centre lies on x -axis and passes through the point (2,3).

Answer:

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

Question:13 Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Answer:

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

Question:14 Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Answer:

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

Question:15 Does the point (–2.5, 3.5) lie inside, outside or on the circle x^2 + y^2 = 25 ?

Answer:

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

Question:1 Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

y^2 =12x

Answer:

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 .

Question:2 Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

x^2 = 6y

Answer:

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 .

Question:3 Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

y^2 = -8x

Answer:

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 .

Question:4 Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

x^2 = -16y

Answer:

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 .

Question:5 Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

y^2 = 10x

Answer:

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 .

Question:6 Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

x^2 = -9y

Answer:

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 .

Question:7 Find the equation of the parabola that satisfies the given conditions:

Focus (6,0); directrix x = - 6

Answer:

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 .

Question:8 Find the equation of the parabola that satisfies the given conditions:

Focus (0,–3); directrix y = 3

Answer:

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 .

Question:9 Find the equation of the parabola that satisfies the given conditions:

Vertex (0,0); focus (3,0)

Answer:

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 .

Question:10 Find the equation of the parabola that satisfies the given conditions:

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

Answer:

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 .

Question:11 Find the equation of the parabola that satisfies the given conditions:

Vertex (0,0) passing through (2,3) and axis is along x -axis.

Answer:

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

Question:12 Find the equation of the parabola that satisfies the given conditions:

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

Answer:

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

Question:1 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

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

Answer:

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}

Question:2 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

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

Answer:

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}

Question:3 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

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

Answer:

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}

Question:4 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

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

Answer:

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

Question:5 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

\frac{x^2}{49} + \frac{y^2}{36} = 1

Answer:

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}

Question:6 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

\frac{x^2}{100} + \frac{y^2}{400} =1

Answer:

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

Question:7 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

36x^2 + 4y^2 =144

Answer:

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}

Question:8 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

16x^2 + y^2 = 16

Answer:

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}

Question:9 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

4x^2 + 9y^2 =36

Answer:

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}

Question:10 Find the equation for the ellipse that satisfies the given conditions:

Vertices (± 5, 0), foci (± 4, 0)

Answer:

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 .

Question:11 Find the equation for the ellipse that satisfies the given conditions:

Vertices (0, ± 13), foci (0, ± 5)

Answer:

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 .

Question:12 Find the equation for the ellipse that satisfies the given conditions:

Vertices (± 6, 0), foci (± 4, 0)

Answer:

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 .

Question:13 Find the equation for the ellipse that satisfies the given conditions:

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

Answer:

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 .

Question:14 Find the equation for the ellipse that satisfies the given conditions:

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

Answer:

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 .

Question:15 Find the equation for the ellipse that satisfies the given conditions:

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

Answer:

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 .

Question:16 Find the equation for the ellipse that satisfies the given conditions:

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

Answer:

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 .

Question:17 Find the equation for the ellipse that satisfies the given conditions:

Foci (± 3, 0), a = 4

Answer:

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 .

Question:18 Find the equation for the ellipse that satisfies the given conditions:

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

Answer:

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 .

Question:19 Find the equation for the ellipse that satisfies the given conditions:

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

Answer:

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 .

Question:20 Find the equation for the ellipse that satisfies the given conditions:

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

Answer:

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

Question:1 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

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

Answer:

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}

Question:2 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

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

Answer:

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

Question:3 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

9 y^2 - 4 x^2 =36

Answer:

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

Question:4 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

16x^2 - 9y^2 = 576

Answer:

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}

Question:5 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

5y^2 - 9x^2 = 36

Answer:

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}

Question:6 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

49y^2 - 16x^2 = 784

Answer:

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}

Question:7 Find the equations of the hyperbola satisfying the given conditions.

Vertices (± 2, 0), foci (± 3, 0)

Answer:

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

Question:8 Find the equations of the hyperbola satisfying the given conditions.

Vertices (0, ± 5), foci (0, ± 8)

Answer:

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 .

Question:9 Find the equations of the hyperbola satisfying the given conditions.

Vertices (0, ± 3), foci (0, ± 5)

Answer:

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 .

Question:10 Find the equations of the hyperbola satisfying the given conditions.

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

Answer:

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

Question:11 Find the equations of the hyperbola satisfying the given conditions.

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

Answer:

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 .

Question:12 Find the equations of the hyperbola satisfying the given conditions.

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

Answer:

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

Question:13 Find the equations of the hyperbola satisfying the given conditions.

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

Answer:

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

Question:14 Find the equations of the hyperbola satisfying the given conditions.

vertices (± 7,0), e = \frac{4}{3}

Answer:

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

Question:15 Find the equations of the hyperbola satisfying the given conditions.

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

Answer:

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

Question:1 If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Answer:

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.

Question:2 An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

Answer:

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}.

Question:3 The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m Find the length of a supporting wire attached to the roadway 18 m from the middle

Answer:

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.

Question:4 An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

Answer:

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.

Question:5 A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

Answer:

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

Question:6 Find the area of the triangle formed by the lines joining the vertex of the parabola x ^2 = 12y to the ends of its latus rectum.

Answer:

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.

Question:7 A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Answer:

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

Question:8 An equilateral triangle is inscribed in the parabola y^2 = 4 ax , where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Answer:

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

Maths chapter 11 class 11 Conic Sections - Topics

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 - \(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 - 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 - \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 - \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.

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NCERT Solutions For Class 11 Mathematics

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

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.

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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.

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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)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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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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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
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zero\;

 Option 4)

K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

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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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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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With increase of temperature, which of these changes?

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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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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twice that in 60 g carbon

 Option 2)

6.023 × 1022
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half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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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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less than 3

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more than 3 but less than 6
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more than 6 but less than 9

 Option 4)

more than 9
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