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NCERT Solutions for Class 11 Maths Chapter 11: Conic Sections - Exercise 11.2

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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2 (Ex 11.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 11 Conic Sections Exercise 11.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 11

Subject:

Class 11 Maths

Chapter Name:

Chapter 11 - Conic Sections

Exercise:

Exercise - 11.2

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.2

1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of latus rectum for ${y^2} = 12x$.

Ans: The given equation is ${y^2} = 12x$. The coefficient of $x$ is positive for the given equation. The general equation of such a parabola is ${y^2} = 4ax$ and the coordinates of focus are $\left( {a,0} \right)$.Therefore, the parabola opens towards the right side.

On comparing the equation, ${y^2} = 12x$ with ${y^2} = 4ax$, we get,

$\Rightarrow 4a = 12$

Divide both sides by 4.

$\Rightarrow a = \dfrac{{12}}{4} $

$\Rightarrow a = 3 $ 

Thus, the coordinates of the focus of the parabola, ${y^2} = 12x$ are $\left( {3,0} \right)$.

The given equation is of the form, ${y^2} = 4ax$ so the axis of the parabola is the $x$-axis.

The equation of the directrix will be given by $x =  - a$. Thus, 

$\Rightarrow x =  - 3 $

$\Rightarrow x + 3 = 0 $ 

The length of the latus rectum is given by $4a$. Substitute $a$ as 3 in $4a$.

$\Rightarrow 4a = 4 \times 3 $   

$\Rightarrow 4a = 12 $ 

Hence, the coordinates of the focus are $\left( {3,0} \right)$, the axis is the $x$-axis, the equation of the directrix is $x + 3 = 0$, and the length of the latus rectum 12.


2. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for ${x^2} = 6y$.

Ans: The given equation is ${x^2} = 6y$. The coefficient of $y$ is positive for the given equation. The general equation of the parabola is ${x^2} = 4ay$ and the coordinates of focus are $\left( {0,a} \right)$.Therefore, the parabola faces upwards.

On comparing the equation, ${x^2} = 6y$ with ${x^2} = 4ay$, we get,

$\Rightarrow 4a = 6$

Divide both sides by 4.

$\Rightarrow a = \dfrac{6}{4} $

$\Rightarrow a = \dfrac{3}{2} $ 

Thus, the coordinates of the focus of the parabola, ${x^2} = 6y$ are $\left( {0,\dfrac{3}{2}} \right)$.

The given equation is of the form, ${x^2} = 4ay$ so the axis of the parabola is the $y$-axis.

The equation of the directrix will be given by $y =  - a$. Thus, 

$\Rightarrow y =  - \dfrac{3}{2} $

$\Rightarrow y + \dfrac{3}{2} = 0 $ 

The length of the latus rectum is given by $4a$. Substitute $a$ as $\dfrac{3}{2}$ in $4a$.

$\Rightarrow 4a = 4 \times \dfrac{3}{2} $

$\Rightarrow 4a = 6 $ 

Hence, the coordinates of the focus are $\left( {0,\dfrac{3}{2}} \right)$, the axis is the $y$-axis, the equation of directrix is $y + \dfrac{3}{2} = 0$, and the length of the latus rectum 6.


3. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of latus rectum for ${y^2} =  - 8x$.

Ans: The given equation is ${y^2} =  - 8x$. The coefficient of $x$ is negative for the given equation. The general equation of such a parabola is ${y^2} =  - 4ax$ and the coordinates of focus are $\left( { - a,0} \right)$.Therefore, the parabola will open towards the left side.

On comparing the equation, ${y^2} =  - 8x$ with ${y^2} =  - 4ax$, we get,

$\Rightarrow  - 4a =  - 8$

Divide both sides by 4.

$\Rightarrow  - a = \dfrac{{ - 8}}{4} $

$\Rightarrow  - a =  - 2 $

$\Rightarrow a = 2 $ 

Thus, the coordinates of the focus of the parabola, ${y^2} =  - 8x$ are $\left( { - 2,0} \right)$.

The given equation is of the form, ${y^2} =  - 4ax$ so the axis of the parabola is the $x$-axis.

The equation of the directrix will be given by $x = a$. Thus, 

$\Rightarrow x = 2 $

$\Rightarrow x - 2 = 0 $ 

The length of the latus rectum is given by $4a$. Substitute $a$ as 2 in $4a$.

$\Rightarrow 4a = 4 \times 2 $

$\Rightarrow 4a = 8 $ 

Hence, the coordinates of the focus are $\left( { - 2,0} \right)$, the axis is the $x$-axis, the equation of directrix is $x - 2 = 0$, and the length of the latus rectum 8.


4. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for ${x^2} =  - 16y$.

Ans: The given equation is ${x^2} =  - 16y$. The coefficient of $y$ is negative for the given equation. The general equation of parabola is ${x^2} =  - 4ay$ and the coordinates of focus are $\left( {0, - a} \right)$. Therefore, the parabola faces downwards.

On comparing the equation, ${x^2} =  - 16y$ with ${x^2} =  - 4ay$, we get,

$\Rightarrow  - 4a =  - 16$

Divide both sides by 4.

$\Rightarrow  - a = \dfrac{{ - 16}}{4} $

$\Rightarrow  - a =  - 4 $

$\Rightarrow a = 4 $ 

Thus, the coordinates of the focus of the parabola, ${x^2} =  - 16y$ are $\left( {0, - 4} \right)$.

The given equation is of the form, ${x^2} =  - 4ay$ so the axis of the parabola is the $y$-axis.

The equation of the directrix will be given by $y = a$. Thus, 

$\Rightarrow y = 4 $

$\Rightarrow y - 4 = 0 $ 

The length of the latus rectum is given by $4a$. Substitute $a$ as 4 in $4a$.

$\Rightarrow 4a = 4 \times 4 $

$\Rightarrow 4a = 16 $ 

Hence, the coordinates of the focus are $\left( {0, - 4} \right)$, the axis is the $y$-axis, the equation of directrix is $y - 4 = 0$, and the length of the latus rectum 16.


5. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of latus rectum for ${y^2} = 10x$.

Ans: The given equation is ${y^2} = 10x$. The coefficient of $x$ is positive for the given equation. The general equation of such a parabola is ${y^2} = 4ax$ and the coordinates of focus are $\left( {a,0} \right)$. Therefore, the parabola opens towards the right side.

On comparing the equation, ${y^2} = 10x$ with ${y^2} = 4ax$, we get,

$\Rightarrow 4a = 10$

Divide both sides by 4.

$\Rightarrow a = \dfrac{{10}}{4} $

$\Rightarrow a = \dfrac{5}{2} $ 

Thus, the coordinates of the focus of the parabola, ${y^2} = 10x$ are $\left( {\dfrac{5}{2},0} \right)$.

The given equation is of the form, ${y^2} = 4ax$ so the axis of the parabola is the $x$-axis.

The equation of the directrix will be given by $x =  - a$. Thus, 

$\Rightarrow x =  - \dfrac{5}{2} $

$\Rightarrow x + \dfrac{5}{2} = 0 $ 

The length of the latus rectum is given by $4a$. Substitute $a$ as $\dfrac{5}{2}$ in $4a$.

$\Rightarrow 4a = 4 \times \dfrac{5}{2} $

$\Rightarrow 4a = 2 \times 5 $

$\Rightarrow 4a = 10 $ 

Hence, the coordinates of the focus are $\left( {\dfrac{5}{2},0} \right)$, the axis is the $x$-axis, the equation of directrix is $x + \dfrac{5}{2} = 0$, and the length of the latus rectum 10.


6. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for ${x^2} =  - 9y$.

Ans: The given equation is ${x^2} =  - 9y$. The coefficient of $y$ is negative for the given equation. The general equation of parabola is ${x^2} =  - 4ay$ and the coordinates of focus are $\left( {0, - a} \right)$. Therefore, the parabola faces downwards.

On comparing the equation, ${x^2} =  - 9y$ with ${x^2} =  - 4ay$, we get,

$\Rightarrow  - 4a =  - 9$

Divide both sides by 4.

$\Rightarrow  - a = \dfrac{{ - 9}}{4} $

$\Rightarrow a = \dfrac{9}{4} $ 

Thus, the coordinates of the focus of the parabola, ${x^2} =  - 9y$are $\left( {0, - \dfrac{9}{4}} \right)$.

The given equation is of the form, ${x^2} =  - 4ay$ so the axis of the parabola is the $y$-axis.

The equation of the directrix will be given by $y = a$. Thus, 

$\Rightarrow y = \dfrac{9}{4}$

$\Rightarrow y - \dfrac{9}{4} = 0$ 

The length of the latus rectum is given by $4a$. Substitute $a$ as $\dfrac{9}{4}$ in $4a$.

$\Rightarrow 4a = 4 \times \dfrac{9}{4} $

$\Rightarrow 4a = 9 $ 

Hence, the coordinates of the focus are $\left( {0, - \dfrac{9}{4}} \right)$, the axis is the $y$-axis, the equation of directrix is $y - \dfrac{9}{4} = 0$, and the length of the latus rectum 9.


7. Find the equation of the parabola that satisfies the following condition: Focus $\left( {6,0} \right)$; directrix $x =  - 6$.

Ans: The given focus is $\left( {6,0} \right)$. We can observe from the coordinates that the focus lies on the $x$-axis. This implies that the axis of the parabola is $x$-axis. Now, from all these observations we can conclude that the parabola is either of the form ${y^2} = 4ax$ or ${y^2} =  - 4ax$. 

The directrix is $x =  - 6$, that lies in the left of the $y$ axis, whereas see that the focus is $\left( {6,0} \right)$, which lies to the right of the $x$ axis.

Hence, the parabola will be of the form, ${y^2} = 4ax$.

Since the coordinates of the focus of a parabola of the form, ${y^2} = 4ax$ are $\left( {a,0} \right)$, then, $a = 6$.

Substitute $a$ as 6 in the equation ${y^2} = 4ax$ to get the equation.

$\Rightarrow {y^2} = 4\left( 6 \right)x $

$\Rightarrow {y^2} = 24x $

Hence, the required equation is \[{y^2} = 24x\].


8. Find the equation of the parabola that satisfies the following condition: Focus $\left( {0, - 3} \right)$; directrix $y = 3$.

Ans: The given focus is $\left( {0, - 3} \right)$. We can observe from the coordinates that the focus lies on the $y$-axis. This implies that the axis of the parabola is $y$-axis. Now, from all these observations we can conclude that the parabola is either of the form ${x^2} = 4ay$or ${x^2} =  - 4ay$. 

The directrix is $y = 3$, that lies above the $x$ axis, whereas see that the focus is $\left( {0, - 3} \right)$, which lies below the$x$ axis. Hence, the parabola will be of the form, ${x^2} =  - 4ay$.

Since the coordinates of the focus of a parabola of the form, ${x^2} =  - 4ay$ are $\left( {0, - a} \right)$, then, $a = 3$.

Substitute $a$ as 3 in the equation ${x^2} =  - 4ay$ to get the equation.

$\Rightarrow {x^2} =  - 4\left( 3 \right)y $

$\Rightarrow {x^2} =  - 12y$

Hence, the required equation is \[{x^2} =  - 12y\].


9. Find the equation of the parabola that satisfies the following condition: Vertex $\left( {0,0} \right)$; focus $\left( {3,0} \right)$.

Ans: The coordinates of the focus are $\left( {3,0} \right)$. This implies that the focus lies on the positive $x$-axis. Thus, $x$-axis is the axis of the parabola. From these observations, the equation of the parabola is of the form ${y^2} = 4ax$.

Since the coordinates of the focus of a parabola of the form, ${y^2} = 4ax$ are $\left( {a,0} \right)$, then, for the given parabola$a = 3$.

Substitute $a$ as 3 in the equation ${y^2} = 4ax$ to get the equation.

$\Rightarrow {y^2} = 4\left( 3 \right)x $

$\Rightarrow {y^2} = 12x$

Hence, the required equation is \[{y^2} = 12x\].


10. Find the equation of the parabola that satisfies the following condition: Vertex $\left( {0,0} \right)$; focus $\left( { - 2,0} \right)$.

Ans: The coordinates of the focus are $\left( { - 2,0} \right)$. This implies that the focus lies on the negative$x$-axis. Thus, $x$-axis is the axis of the parabola. From these observations, the equation of the parabola is of the form ${y^2} =  - 4ax$.

Since the coordinates of the focus of a parabola of the form, ${y^2} =  - 4ax$ are $\left( { - a,0} \right)$, then, for the given parabola $a = 2$.

Substitute $a$ as 2 in the equation ${y^2} =  - 4ax$ to get the equation.

$\Rightarrow {y^2} =  - 4\left( 2 \right)x $

$\Rightarrow {y^2} =  - 8x $ 

Hence, the required equation is ${y^2} =  - 8x$.


11. Find the equation of the parabola that satisfies the following: Vertex $\left( {0,0} \right)$, passing through $\left( {2,3} \right)$ and axis is along $x$-axis.

Ans: The vertex is $\left( {0,0} \right)$. We can observe from the coordinates that the focus lies on the $x$-axis. This implies that the axis of the parabola is $x$-axis. Now, from all these observations we can conclude that the parabola is either of the form ${y^2} = 4ax$or ${y^2} =  - 4ax$. 

The parabola passes through $\left( {2,3} \right)$. These coordinates lie in the first quadrant.

The equation of the parabola will be, ${y^2} = 4ax$ so, the point $\left( {2,3} \right)$ must satisfy it.

Substitute $x$ as 2, $y$ as 3 in ${y^2} = 4ax$ and solve for $a$.

$\Rightarrow {\left( 3 \right)^2} = 4a\left( 2 \right) $

$\Rightarrow 9 = 8a$

Divide both sides by 8.

$\Rightarrow \dfrac{9}{8} = a$

Now, substitute $a$ as $\dfrac{9}{8}$ in ${y^2} = 4ax$ to get the required equation.

$\Rightarrow {y^2} = 4 \times \dfrac{9}{8} \times x $

$\Rightarrow {y^2} = \dfrac{9}{2}x $

$\Rightarrow 2{y^2} = 9x $ 

The equation of the parabola with vertex $\left( {0,0} \right)$ passing through $\left( {2,3} \right)$ and axis is along $x$-axis is $2{y^2} = 9x$.


12. Find the equation of the parabola that satisfies the following conditions: Vertex $\left( {0,0} \right)$, passing through $\left( {5,2} \right)$ and symmetric with respect to $y$-axis.

Ans: The vertex is $\left( {0,0} \right)$ and the parabola is symmetric about the $y$ axis. Wecan conclude that the parabola is either of the form ${x^2} = 4ay$ or ${x^2} =  - 4ay$. 

The parabola passes through $\left( {5,2} \right)$. These coordinates lie in the first quadrant.

The equation of the parabola will be, ${x^2} = 4ay$ so, the point $\left( {5,2} \right)$ must satisfy it.

Substitute $x$ as 5, $y$ as 2 in ${x^2} = 4ay$ and solve for $a$.

$\Rightarrow {\left( 5 \right)^2} = 4 \times a \times \left( 2 \right) $

$\Rightarrow 25 = 8a $ 

Divide both sides by 8.

$\Rightarrow \dfrac{{25}}{8} = a$

Now, substitute $a$ as $\dfrac{{25}}{8}$ in ${x^2} = 4ax$ to get the required equation.

$\Rightarrow {x^2} = 4 \times \dfrac{{25}}{8} \times y $

$\Rightarrow {x^2} = \dfrac{{25}}{2}y $

$\Rightarrow 2{x^2} = 25y $ 

The equation of the parabola with vertex $\left( {0,0} \right)$ passing through $\left( {5,2} \right)$ and symmetric with respect to $x$-axis is $2{x^2} = 25y$.


NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.2

Opting for the NCERT solutions for Ex 11.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 11 Exercise 11.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 11 Exercise 11.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 11 Exercise 11.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well. 

FAQs on NCERT Solutions for Class 11 Maths Chapter 11: Conic Sections - Exercise 11.2

1. Where can I download accurate NCERT Solutions for Class 11 Maths Chapter 11 Conic sections (Ex 11.2) Exercise 11.2?

There are many online platforms where these study materials are available. But you can refer to Vedantu, India’s no. 1 online educational platform. Here NCERT solutions, revision notes, and many other essential study materials for all subjects are designed by top subject experts. All these solutions are given in a detailed step-by-step manner and are 100% accurate. You can download the NCERT Solutions for Class 11 Maths Chapter 11 Conic sections (Ex 11.2) Exercise 11.2 in the free PDF format on the website Vedantu.com or you can download Vedantu mobile app.

2. How can NCERT Solutions for Class 11 Maths Chapter 11 Conic sections (Ex 11.2) Exercise 11.2 be helpful in the final exams?

Conic sections in chapter 11 Class 11 Maths deals with the different shapes, circular cones, etc. While solving these NCERT exercise solutions students may struggle with basic problems. Vedantu provides NCERT solutions without any charge which follow an easy approach so that students get to understand the basics while solving these questions. Apart from that, they will also be prepared for tricky questions before their final exams.

3. How many topics are there in Chapter 11 Conic sections of NCERT Class 11 Maths?

There are many topics and subtopics in Chapter 11 Conic sections and these are listed below:

11.1)Introduction

11.2)Section of a Cone

11.2.1)Ellipse, Circle, Parabola, Hyperbola

11.2.2)Degenerate Conic Sections

11.3)Circle

11.4)Parabola

11.4.1)Standard Equation of a Parabola

11.4.2)Latus Rectum

11.5)Ellipse

11.6)Hyperbola

11.6.1)Eccentricity

11.6.2)Standard Equation of a Parabola

11.6.3)Latus Rectum

4. Is chapter 11 conic sections in NCERT Class 11 Maths important?

Absolutely yes, Chapter 11 Conic sections are a very important chapter in Class 11 Maths. This will help you with a base for the upcoming classes and also be useful for any Competitive exams. So, students should give extra focus on this chapter and understand the concepts. You can refer to Vedantu, where you can get detailed and stepwise solutions of Chapter 11 Conic sections in the free PDF format. Also, you will be able to build a better understanding level and learn a different approach to solving the problem of Conic sections.

5. How many questions are there in NCERT Solutions for Class 11 Maths Chapter 11 Conic sections (Ex 11.2) Exercise 11.2?

There are a total of four exercises in chapter 11 of the NCERT Class 11 Maths and exercise 11.2 consists of 12 questions. Solutions to all these questions are being provided by Vedantu and these are available on the official website of Vedantu and the mobile application of Vedantu. The best part is that these NCERT solutions are specially designed by qualified teacher experts and these are available for students free of cost.