NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

Edited By Komal Miglani | Updated on Mar 24, 2025 06:51 PM IST

NCERT Maths Chapter 10 class 11 discusses conic sections and their solutions. This is a continuation of the last chapter of straight lines and here we will discuss mainly four other curves: Circle, Hyperbola, Parabola, and Ellipse. “Conic” means “Cone” and “Sections” means “Intersection”. These curves are called “Conic sections” because they can be formed by the intersection of a plane with a double-napped right circular cone. Now, strengthening the basic concepts is an absolute necessity to solve the exercises in this chapter.

This Story also Contains
  1. Conic Sections Class 11 Questions And Answers PDF Free Download
  2. Conic Sections Class 11 Solutions: Important Formulae
  3. Conic Sections Class 11 NCERT Solutions (Exercise)
  4. Importance of solving NCERT Questions for Class 11 Chapter 10 Conic Sections
  5. NCERT Solutions For Class 11: Subject Wise
  6. 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

This article will explain all the conic sections class 11 questions and answers based on the latest NCERT 2025-26 syllabus in detail. Conic sections NCERT solutions are not only useful for CBSE board exams but also for competitive exams like JEE Mains, JEE Advanced, VITEEE, and BITSAT. Learnings from this chapter can be used in real-life applications like Wheels, satellite dishes, headlights, GPS, planetary orbits, MRI machines and many more. For a quick revision, students can use the PDFs of class 11 Maths chapter 10 NCERT solutions. Conic Section Class 11 Maths Chapter Notes and NCERT Exemplar Solutions For Class 11 Maths Chapter Conic Sections can also be used for mastering this chapter and deeper knowledge gain purposes.

Conic Sections Class 11 Questions And Answers PDF Free Download

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Conic Sections Class 11 Solutions: Important Formulae

Circle

Description

Equation/Formulae

Equation of a circle

(x - h)2 + (y - k)2 = r2

General equation of a circle

x2 + y2 + 2gx + 2fy + c = 0

Centre of the circle

Centre: (-g, -f)

Radius of the circle

Radius (r) = $\sqrt{g^2+f^2-c}$

Parametric equation of a circle

x = r cos(θ), y = r sin(θ)

Parametric equation of a circle (centre 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 the latus rectum

4a (for all forms)



Ellipse

Description

Equation/Forms

Equation forms of ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b$
And
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1(a>b)$

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 = $\pm \frac ae$ (1st form), y = $\pm \frac ae$ (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 the latus rectum

$\frac{2b^2}a$ (for both forms)

Eccentricity (e)

$\sqrt{\frac{a^2+b^2}{a^2}}$(for both forms)


Hyperbola

Description

Equations/Forms

Equation forms of the hyperbola

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

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 = $\pm\frac ae$ (for both forms)

Eccentricity (e)

$\sqrt{\frac{(a^2+b^2)}{a^2}}$(for both forms)

Length of the latus rectum

$\frac{2b^2}a$ (for both forms)



























































Conic Sections Class 11 NCERT Solutions (Exercise)

NCERT Conic Sections Class 11 Solutions: Exercise: 11.1
Page Number: 181
Total Questions: 15


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:

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

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:

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

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 centre ( h, k) and radius r is given by:

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

Here, $(h,k)=\left ( \frac{1}{2},\frac{1}{4} \right )$ and $r=\frac{1}{12}$

So the equation of the 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:

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

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:

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

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:

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

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

It can also be written in the form: $(x-(-5))^2 + (y-3)^2 = 6^2$
After 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:

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

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

It can also be written in the form: $(x-2)^2 + (y-4)^2 =(\sqrt{65})^2$

After 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:

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

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

It can also be written in the form: $(x-4)^2 + (y-(-5))^2 = (\sqrt{53})^2$

After 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:

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

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

It can also be written in the form: $\left ( x-\frac{1}{4}\right )^2 + \left ( y-0 \right )^2 = \left ( \frac{1}{4} \right )^2$

After comparing, we can see that,

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

Hence, centre of the circle is the $\left ( \frac{1}{4},0\right )$ and 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:

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

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,

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

$⇒(4-h)^2-(6-h)^2+(1-k)^2-(5-k)^2=0$

$⇒(-2)(10-2h)+(-4)(6-2k)=0$

$⇒-20+4h-24+8k=0$

$⇒4h+8k=44$

Condition 2: Centre is on the line $4x + y = 16$.

So, $4h+k=16$

From condition 1 and condition 2, we get,

$h=3$ and $k=4$

Substituting the values of h and k in condition 1, we get,

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

$⇒1+9=r^2$

$⇒r=\sqrt{10}$

$\therefore$ 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 whose 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 ;

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

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,

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

$⇒(2-h)^2-(-1-h)^2+(3-k)^2-(1-k)^2=0$

$⇒(3)(1-2h)+(2)(4-2k)=0$

$⇒3-6h+8-4k=0$

$⇒6h+4k=11$

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

So, $h-3k=11$

From condition 1 and condition 2, we get,

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

Substituting the values of h and k in condition 1, we get,

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

$⇒\frac{9}{4}+\frac{121}{4}=r^2$

$⇒r^2=\frac{130}{4}$

$\therefore$ 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, $h=6$

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, the center of the circle is (0, 0).

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

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

$⇒d=\sqrt{6.25+12.25}$

$⇒d=\sqrt{18.5}\approx 4.3<5$

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

NCERT Conic Sections Class 11 Solutions: Exercise: 11.2
Page Number: 186-187
Total Questions: 12

Question:1 Find the coordinates of the focus, the 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 a 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, the 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 a 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, the 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 a parabola of the form $y^2=-4ax$ which opens towards the 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, the 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 a 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:

$4a=4\times4=16$

Question:5 Find the coordinates of the focus, the 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 a 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:

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

Question:6 Find the coordinates of the focus, the axis of the parabola, the equation of the directrix and the length of the latus rectum: $x^2 = -9y$

Answer:

Given, a parabola with an equation $x^2 =-9y$

This is a 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.
Directrix is of the form $x=-a$ which means it lies left to the Y-Axis.

These are the conditions 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.
Directrix is of the form $y=a$ which means it lies above the 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 the vertex of the parabola is (0,0) and the focus lies on the positive X-axis, The parabola will open towards the right, And the standard equation of such a 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 the vertex of the parabola is (0,0) and the focus lies in the negative X-axis, The parabola will open towards the left, And the standard equation of such a 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 the x-axis.

Answer:

Given

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

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

$y^2=4ax$

Now since it passes through (2,3)

$3^2=4a(2)$

$⇒9=8a$

$⇒a=\frac{9}{8}$

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 the 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 the Y-axis, its axis will be Y-axis. and since it passes through the point (5,2), it must go through the first quadrant.

So the standard equation of such a 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 Conic Sections Class 11 Solutions: Exercise: 11.3
Page Number: 195
Total Questions: 20

Question:1 Find the coordinates of the foci, the vertices, the length of the 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:

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 the X-axis and the minor axis is along the 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,

$c=\sqrt{a^2-b^2}=\sqrt{6^2-4^2}=\sqrt{20}=2\sqrt5$

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 the 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 the 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:

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 the Y-axis and the minor axis is along the X-axis.

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

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

We get,

$a=5$ and $b=2$

So,

$c=\sqrt{a^2-b^2}=\sqrt{5^2-2^2}=\sqrt{21}$

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 the 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 the 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:

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 the X-axis and the minor axis is along the 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,

$c=\sqrt{a^2-b^2}=\sqrt{4^2-3^2}=\sqrt7$

Hence, the 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 the 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 the 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:

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 the Y-axis and the minor axis is along the X-axis.

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

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

We get,

$a=10$ and $b=5$

So,

$c=\sqrt{a^2-b^2}=\sqrt{10^2-5^2}=\sqrt{75}=5\sqrt3$

Hence, the 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 the 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 the 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:

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 the X-axis and the minor axis is along the 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,

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

So,

$c=\sqrt{a^2-b^2}=\sqrt{7^2-6^2}=\sqrt{13}$

Hence, the 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 the major axis:

$2a=2(7)=14$

The length of the 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 the 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:

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 the Y-axis and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an 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 the 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 the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: $36x^2 + 4y^2 =144$

Answer:

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 the Y-axis and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an 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, the 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 the 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 the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: $16x^2 + y^2 = 16$

Answer:

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 the Y-axis and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an 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, the 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 the 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 the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: $4x^2 + 9y^2 =36$

Answer:

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 the X-axis and the minor axis is along the 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, the 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 the 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 :

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

$⇒\frac{x^2}{25}+\frac{y^2}{9}=1$ .

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 :

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

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

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 :

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

$⇒\frac{x^2}{36}+\frac{y^2}{20}=1$ .

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 the 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 :

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

$⇒\frac{x^2}{9}+\frac{y^2}{4}=1$ .

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 the 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 :

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

$⇒\frac{x^2}{1}+\frac{y^2}{5}=1$ .

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 :

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

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

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 :

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

$⇒\frac{x^2}{64}+\frac{y^2}{100}=1$ .

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 :

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

$⇒\frac{x^2}{16}+\frac{y^2}{7}=1$ .

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 :

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

$⇒\frac{x^2}{25}+\frac{y^2}{9}=1$ .

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 equations, 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 is 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 these two equations, 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 Conic Sections Class 11 Solutions: Exercise: 11.4
Page Number: 202
Total Questions: 15

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

It 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, we get,

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

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 the axis of the hyperbola is the 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, $\frac{y^2}{9} - \frac{x^2}{27} = 1$

It 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 the axis of the hyperbola is the 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, $9 y^2 - 4 x^2 =36$

It 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, the 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, $16x^2 - 9y^2 = 576$

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

It 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 the axis of the hyperbola is the 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, $49y^2 - 16x^2 = 784$

It 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 foci 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 foci 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 foci 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, foci 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, foci 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, foci 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 (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, $a=5$

Since $a$ can never be negative,

$a=5$

$⇒a^2=25$

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

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, foci 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 (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, $a=2$

Since $a$ can never be negative,

$a=2$

$⇒a^2=4$

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

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

$e=\frac{c}{a}$
$⇒\frac{4}{3}=\frac{c}{7}$
$⇒c=\frac{28}{3}$

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 the foci of the hyperbola are on the 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 the 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), we get,

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

$⇒a^4-23a^2+90=0$

$⇒(a^2)^2-18a^2-5a^2+90=0$

$⇒(a^2-18)(a^2-5)=0$

$⇒a^2=18$ or, $a=5$

Now, as we know in a hyperbola $c$ is always greater than $a$,
we choose the value $a^2=5$

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

Hence The Equation of the hyperbola is $\frac{y^2}{5}-\frac{x^2}{5}=1$.

NCERT Conic Sections Class 11 Solutions: Exercise: Miscellaneous Exercise
Page Number: 204
Total Questions: 8

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

Answer:

Let the parabolic reflector open towards the right.

So, the equation of the 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, focus, $f=\frac{R}{2}$

$\therefore$ 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 ($\frac52$, 10) if we assume the 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 upwards, the equation will be of the form $x^2=4ay$

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

$50^2=4a\times 24$

$⇒a=\frac{625}{24}$

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:

$y=\frac{x^2}{4a}=\frac{18^2}{4\times \frac{625}{24}}\approx 3.11$ m

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

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 semi-major 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

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

$⇒y\approx 1.56$ m

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 the rod makes with the ground.

Now, at a point 3 cm from the end,

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

At the point of touching the ground

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

Now, As we know the trigonometric identity,

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

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

$⇒\frac{x^2}{81}+\frac{y^2}{9}=1$

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 the ends of the latus rectum are:

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

So, the coordinates of the 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)

Width of the triangle $= 2 \times 6=12$

Height of the triangle = 3

So, the area of the triangle
= $\frac{1}{2}\times$ Base $\times$ height
= $\frac{1}{2}\times12\times3$
= $18$

Hence, the required area is 18 square units.

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 if a point moves in a plane in such a way that its distance from two points remains 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:

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

Here, students can check the solved examples before moving on to the exercises. Exercises are also prepared based on difficulty levels varying from easy to moderate level. These links are separate links to the exercises. Students can check each exercise separately and try to solve the next one on their own before checking the solutions.

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Importance of solving NCERT Questions for Class 11 Chapter 10 Conic Sections

Class 11 maths chapter 10 question answers are essential for many things. But before solving them, strengthen your fundamentals of conic sections. Some importance of solving these problems are listed below.

  • A healthy amount of questions come from this chapter in the final exams. So by mastering this chapter, students can achieve higher grades.
  • This chapter is also the backbone of some other mathematical and science topics like coordinate geometry, calculus, and physics applications.
  • Questions from this chapter appear in important exams like JEE mains, VITEE, and BITSAT. So, studying them beforehand helps to reduce the pressure of facing questions from this chapter during these exams.
  • The diagrams of conic sections help visualize the various parts of these shapes and symmetry. This will be helpful for further studies like engineering, architecture, and data science.

All in all, mastering this chapter is an absolute necessity for students in class 11.

NCERT Solutions For Class 11: Subject Wise

Students can use the following links to check the solutions of the other subjects to achieve good marks in the exam.

NCERT Books and NCERT Syllabus

After studying the NCERT books for fundamental knowledge, students can solve problems from other books. Also, they can check the latest NCERT syllabus to stay updated using the following links.

Frequently Asked Questions (FAQs)

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. 

Articles

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

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

 

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

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\;

Option 3)

zero\;

Option 4)

K/4

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 .

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

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.

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.

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

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

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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