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# NCERT Solutions Class 11 Maths Chapter 10 Conic Sections

Last updated date: 20th Sep 2024
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## NCERT Solutions for Maths Chapter 10 Conic Sections Class 11 - FREE PDF Download

Attaining full marks in Mathematics should not be that difficult if you have access to the right kind of study material. NCERT Class 11 Maths Chapter 10 comprises an interesting chapter called Conic Sections that include descriptions of Ellipses, Parabolas, Circles, and Hyperbolas. These concepts may seem a little difficult to grasp in the beginning, since there are tons of equations and tricky methods. However, Conic Sections Class 11 Solutions perfectly assist in dealing with these problems.

Table of Content
1. NCERT Solutions for Maths Chapter 10 Conic Sections Class 11 - FREE PDF Download
2. Glance on Maths Chapter 10 Class 11 - Conic Sections
3. Access Exercise Wise NCERT Solutions for Chapter 10 Maths Class 11
4. Exercises Under NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections
5. Access NCERT Solutions for Class 11 Maths Chapter 10 – Conic Sections
5.1Exercise  10.1
5.2Exercise 10.2
5.3Exercise 10.3
5.4Exercise 10.4
5.5Miscellaneous Exercise
6. Summary of Conic Sections
7. Overview of Deleted Syllabus for CBSE Class 11 Maths Chapter 10 Conic Sections
8. Class 11 Maths Chapter 10: Exercises Breakdown
9. Other Study Material for CBSE Class 11 Maths Chapter 10
10. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

Studying the chapters only is never enough to master a subject like Mathematics; hence, students ought to practise exercise questions along with that. Class 11 Maths NCERT Solutions are prepared by experienced professionals to help you get accustomed to uncommon problems.

## Glance on Maths Chapter 10 Class 11 - Conic Sections

• Conic Sections Class 11 explains the different types of conic sections and their properties.

• A conic section is a curve obtained by intersecting a cone with a plane.

• An ellipse is the set of all points in a plane where the sum of the distances from two fixed points is constant.

• A hyperbola is the set of all points in a plane where the difference of the distances from two fixed points is constant.

• A parabola is the set of all points in a plane that are equidistant from a fixed point and a fixed line.

• The chapter covers the derivation of standard equations for each conic section.

• This article contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter 10 - Conic Sections Class 11, which you can download as PDFs.

• There are five exercises including Miscellaneous Exercise (70 fully solved questions) in class 11th maths chapter 10 Conic Sections.

## Access Exercise Wise NCERT Solutions for Chapter 10 Maths Class 11

 S.No. Current Syllabus Exercises of Class 11 Maths Chapter 10 1 NCERT Solutions of Class 11 Maths Conic Sections Exercise 10.1 2 NCERT Solutions of Class 11 Maths Conic Sections Exercise 10.2 3 NCERT Solutions of Class 11 Maths Conic Sections Exercise 10.3 4 NCERT Solutions of Class 11 Maths Conic Sections Exercise 10.4
Competitive Exams after 12th Science

## Access NCERT Solutions for Class 11 Maths Chapter 10 – Conic Sections

### Exercise  10.1

1. Find the equation of the circle with centre $(0,2)$ and radius $2$

Ans: The equation of a circle with centre $(h,k)$ and radius $r$ is given as

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

It is given that centre $(h,k)=(0,2)$ and radius $(r)=2.$

Therefore, the equation of the circle is

${{(x-0)}^{2}}+{{(y-2)}^{2}}={{2}^{2}}$

\begin{align} & {{x}^{2}}+{{y}^{2}}+4-4y=4 \\ & {{x}^{2}}+{{y}^{2}}-4y=0 \\ \end{align}

2. Find the equation of the circle with centre $(-2,3)$ and radius $4$

Ans: The equation of a circle with centre $(h,k)$ and radius $r$ is given as

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

It is given that centre $(h,k)=(-2,3)$ and radius $(r)=4.$

Therefore, the equation of the circle is

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

\begin{align} & {{x}^{2}}+4x+4+{{y}^{2}}-6y+9=16 \\ & {{x}^{2}}+{{y}^{2}}+4x-6y-3=0 \\ \end{align}

3. Find the equation of the circle with centre $\left( \dfrac{1}{2},\dfrac{1}{4} \right)$ and radius $\left( \dfrac{1}{12} \right)$

Ans: The equation of a circle with centre $(h,k)$ and radius $r$ is given as

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

It is given that centre $(h,k)=\left( \dfrac{1}{2},\dfrac{1}{4} \right)$ and radius $(r)=\left( \dfrac{1}{12} \right).$

Therefore, the equation of the circle is

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

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

${{x}^{2}}-x+\dfrac{1}{4}+{{y}^{2}}-\dfrac{y}{2}+\dfrac{1}{16}-\dfrac{1}{144}=0$

$144{{x}^{2}}-144x+36+144{{y}^{2}}-72y+9-1=0$ (Solve by taking LCM)

$144{{x}^{2}}-144x+144{{y}^{2}}-72y+44=0$

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

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

4. Find the equation of the circle with centre $(1,1)$ and radius $\sqrt{2}$

Ans: The equation of a circle with centre $(h,k)$ and radius $r$ is given as

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

It is given that centre $(h,k)=(1,1)$ and radius $(r)=\sqrt{2}.$

Therefore, the equation of the circle is

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

\begin{align} & {{x}^{2}}-2x+1+{{y}^{2}}-2y+1=2 \\ & {{x}^{2}}+{{y}^{2}}-2x-2y=0 \\ \end{align}

5. Find the equation of the circle with centre $(-a,-b)$ and radius $\sqrt{{{a}^{2}}-{{b}^{2}}}$

Ans: The equation of a circle with centre $(h,k)$ and radius $r$ is given as

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

It is given that centre $(h,k)=(-a,-b)$ and radius $(r)=\sqrt{{{a}^{2}}-{{b}^{2}}}.$

Therefore, the equation of the circle is

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

\begin{align} & {{x}^{2}}+2ax+{{a}^{2}}+{{y}^{2}}+2by+{{b}^{2}}={{a}^{2}}-{{b}^{2}} \\ & {{x}^{2}}+{{y}^{2}}+2ax+2by+2{{b}^{2}}=0 \\ \end{align}

6. Find the centre and radius of the circle ${{(x+5)}^{2}}+{{(y-3)}^{2}}=36$

Ans: The equation of the given circle is ${{(x+5)}^{2}}+{{(y-3)}^{2}}=36$

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

$\Rightarrow {{\{x-(-5)\}}^{2}}+{{(y-3)}^{2}}={{(6)}^{2}}$, which is the form of ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h=-5,k=3,and\; r=6$.

Thus, the centre of the given circle is $(-5,3),$ while its radius is 6.

7. Find the centre and radius of the circle ${{x}^{2}}+{{y}^{2}}-4x-8y-45=0$

Ans: The equation of the given circle is ${{x}^{2}}+{{y}^{2}}-4x-8y-45=0$

$\Rightarrow {{x}^{2}}+{{y}^{2}}-4x-8y-45=0$

$\Rightarrow ({{x}^{2}}-4x)+({{y}^{2}}-8y)=45$

$\Rightarrow \{{{x}^{2}}-2(x)(2)+{{(2)}^{2}}\}+\{{{y}^{2}}-2(y)(4)+{{(4)}^{2}}\}-4-16=45$

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

$\Rightarrow {{(x-2)}^{2}}+{{(y-4)}^{2}}={{\left( \sqrt{65} \right)}^{2}},$ which is of the form ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h=2,k=4,and\; r=\sqrt{65}$

Thus, the centre of the given circle is $(2,4),$while its radius is $\sqrt{65}$.

8. Find the centre and radius of the circle ${{x}^{2}}+{{y}^{2}}-8x+10y-12=0$

Ans: The equation of the given circle is ${{x}^{2}}+{{y}^{2}}-8x+10y-12=0$

$\Rightarrow {{x}^{2}}+{{y}^{2}}-8x+10y-12=0$
$\Rightarrow ({{x}^{2}}-8x)+({{y}^{2}}+10y)=12$

$\Rightarrow \{{{x}^{2}}-2(x)(4)+{{(4)}^{2}}\}+\{{{y}^{2}}+2(y)(5)+{{(5)}^{2}}-16-25=12$

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

$\Rightarrow {{(x-4)}^{2}}+{{\{y-(-5)\}}^{2}}={{\left( \sqrt{53} \right)}^{2}},$ which is of the form ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h=4,k=-5,and\; r=\sqrt{53}$

Thus, the centre of the given circle is $(4,-5),$ while its radius is $\sqrt{53}$.

9. Find the centre and radius of the circle $2{{x}^{2}}+2{{y}^{2}}-x=0$

Ans: The equation of the given circle is $2{{x}^{2}}+2{{y}^{2}}-x=0$

$\Rightarrow 2{{x}^{2}}+2{{y}^{2}}-x=0$
$\Rightarrow (2{{x}^{2}}-x)+2{{y}^{2}}=0$

$\Rightarrow 2\left[ \left( {{x}^{2}}-\dfrac{x}{2} \right)+{{y}^{2}} \right]=0$

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

$\Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( \dfrac{1}{4} \right)}^{2}},$ which is of the form ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h=\dfrac{1}{4},k=0,$ and $r = \dfrac{1}{4}$

Thus, the centre of the given circle is $\left( \dfrac{1}{4},0 \right),$ while its radius is $\dfrac{1}{4}$.

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

Ans: Let the equation of the required circle be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Since the circle passes through points $(4,1)$ and $(6,5)$,

${{(4-h)}^{2}}+{{(1-k)}^{2}}={{r}^{2}}$ ………(i)

${{(6-h)}^{2}}+{{(5-k)}^{2}}={{r}^{2}}$ ………(ii)

Since the centre (h,k) of the circle lies on line $4x+y=16,$

$4h+k=16$ ………(iii)

From equations (i) and (ii), we get

$\Rightarrow {{(4-h)}^{2}}+{{(1-k)}^{2}}={{(6-h)}^{2}}+{{(5-k)}^{2}}$

$\Rightarrow 16-8h+{{h}^{2}}+1-2k+{{k}^{2}}=36-12h+{{h}^{2}}+25-10k+{{k}^{2}}$

\begin{align} & \Rightarrow 16-8h+1-2k=36-12h+25-10k \\ & \Rightarrow 4h+8k=44 \\ \end{align}

$\Rightarrow h+2k=11$ ………(iv)

On solving equations (iii) and (iv), we obtain $h=3$ and $k=4$

On substituting the values of h and k in equation (i), we obtain

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

$\Rightarrow {{(1)}^{2}}+{{(-3)}^{2}}={{r}^{2}}$

$\Rightarrow 1+9={{r}^{2}}$

\begin{align} & \Rightarrow {{r}^{2}}=10 \\ & \Rightarrow r=\sqrt{10} \\ \end{align}

Thus, the equation of the required circle is

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

\begin{align} & {{x}^{2}}-6x+9+{{y}^{2}}-8y+16=10 \\ & {{x}^{2}}+{{y}^{2}}-6x-8y+15=0 \\ \end{align}

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

Ans: Let the equation of the required circle be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Since the circle passes through points $(2,3)$ and $(-1,1)$,

${{(2-h)}^{2}}+{{(3-k)}^{2}}={{r}^{2}}$ ………(i)

${{(-1-h)}^{2}}+{{(1-k)}^{2}}={{r}^{2}}$ ………(ii)

Since the centre (h,k) of the circle lies on line $x-3y-11=0$

$h-3k=11$ ………(iii)

From equations (i) and (ii), we get

$\Rightarrow {{(2-h)}^{2}}+{{(3-k)}^{2}}={{(-1-h)}^{2}}+{{(1-k)}^{2}}$

$\Rightarrow 4-4h+{{h}^{2}}+9-6k+{{k}^{2}}=1+2h+{{h}^{2}}+1-2k+{{k}^{2}}$

$\Rightarrow 4-4h+9-6k=1+2h+1-2k$

$\Rightarrow 6h+4k=11$ ………(iv)

On solving equations (iii) and (iv), we obtain $h=\dfrac{7}{2}$ and $k=\dfrac{-5}{2}$

On substituting the values of h and k in equation (i), we obtain

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

$\Rightarrow {{\left( \dfrac{4-7}{2} \right)}^{2}}+{{\left( \dfrac{6+5}{2} \right)}^{2}}={{r}^{2}}$

$\Rightarrow {{\left( \dfrac{-3}{2} \right)}^{2}}+{{\left( \dfrac{11}{2} \right)}^{2}}={{r}^{2}}$

\begin{align} & \Rightarrow \dfrac{9}{4}+\dfrac{121}{4}={{r}^{2}} \\ & \Rightarrow \dfrac{130}{4}={{r}^{2}} \\ \end{align}

Thus, the equation of the required circle is

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

\begin{align} & {{\left( \dfrac{2x-7}{2} \right)}^{2}}+{{\left( \dfrac{2y+5}{2} \right)}^{2}}=\dfrac{130}{4} \\ & 4{{x}^{2}}-28x+49+4{{y}^{2}}+20y+25=130 \\ \end{align}

\begin{align} & 4{{x}^{2}}+4{{y}^{2}}-28x+20y-56=0 \\ & 4({{x}^{2}}+y{}^{2}-7x+5y-14)=0 \\ \end{align}

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

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

Ans: Let the equation of the required circle be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Since the radius of the circle  is 5 and its centre lies on the x-axis, $k=0$ and $r=5$.

Now, the equation of the circle passes through point $(2,3).$

\begin{align} & \therefore {{\left( 2-h \right)}^{2}}+{{3}^{2}}=25 \\ & \Rightarrow {{\left( 2-h \right)}^{2}}=25-9 \\ & \Rightarrow {{\left( 2-h \right)}^{2}}=16 \\ \end{align}

\begin{align} & \Rightarrow \left( 2-h \right)=\pm \sqrt{16} \\ & =\pm 4 \\ \end{align}

If $2-h=4,$ then $h=-2$

If $2-h=-4,$ then $h=6$

When $h=-2$, the equation of the circle becomes

\begin{align} & {{\left( x+2 \right)}^{2}}+{{y}^{2}}=25 \\ & {{x}^{2}}+4x+4+{{y}^{2}}=25 \\ & {{x}^{2}}+{{y}^{2}}+4x-21=0 \\ \end{align}

When $h=6$, the equation of the circle becomes

\begin{align} & {{\left( x-6 \right)}^{2}}+{{y}^{2}}=25 \\ & {{x}^{2}}-12x+36+{{y}^{2}}=25 \\ & {{x}^{2}}+{{y}^{2}}-12x+11=0 \\ \end{align} So, the equation of the circle can be ${{x}^{2}}+{{y}^{2}}+4x-21=0$ or ${{x}^{2}}+{{y}^{2}}-12x+11=0$

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

Ans: Let the equation of the required circle be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Since the circle passes through $(0,0)$.

\begin{align} & {{(0-h)}^{2}}+{{(0-k)}^{2}}={{r}^{2}} \\ & \Rightarrow {{h}^{2}}+{{k}^{2}}={{r}^{2}} \\ \end{align}

The equation of the circle now becomes ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{h}^{2}}+{{k}^{2}}$

It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points $(a,0)$ and $(0,b)$. Therefore,

${{(a-h)}^{2}}+{{(0-k)}^{2}}={{h}^{2}}+{{k}^{2}}$ …….(1)

${{(0-h)}^{2}}+{{(b-k)}^{2}}={{h}^{2}}+{{k}^{2}}$ …….(2)

From equation (1), we’ll get

\begin{align} & {{a}^{2}}-2ah+{{h}^{2}}+{{k}^{2}}={{h}^{2}}+{{k}^{2}} \\ & \Rightarrow {{a}^{2}}-2ah=0 \\ & \Rightarrow a(a-2h)=0 \\ \end{align}

$\Rightarrow a=0$ or $(a-2h)=0$

However, $a\ne 0;$ hence, $(a-2h)=0\Rightarrow h=\dfrac{a}{2}$

From equation (2), we’ll get

\begin{align} & {{h}^{2}}+{{b}^{2}}-2bk+{{k}^{2}}={{h}^{2}}+{{k}^{2}} \\ & \Rightarrow {{b}^{2}}-2bk=0 \\ & \Rightarrow b(b-2k)=0 \\ \end{align}

$\Rightarrow b=0$ or $(b-2k)=0$

However, $b\ne 0;$ hence, $(b-2k)=0\Rightarrow k=\dfrac{b}{2}$

Thus, the equation of the required circle is

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

${{\left( \dfrac{2x-a}{2} \right)}^{2}}+{{\left( \dfrac{2y-b}{2} \right)}^{2}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{4}$

\begin{align} & \Rightarrow 4{{x}^{2}}-4ax+{{a}^{2}}+4{{y}^{2}}-4by+{{b}^{2}}={{a}^{2}}+{{b}^{2}} \\ & \Rightarrow 4{{x}^{2}}+4{{y}^{2}}-4ax-4by=0 \\ & \Rightarrow {{x}^{2}}+{{y}^{2}}-ax-by=0 \\ \end{align}

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

Ans: The centre of the circle is given as $(h,k)=(2,2)$

Since the circle passes through point $(4,5),$ the radius $(r)$ of the circle is the distance between the points $(2,2)$ and $(4,5)$.

\begin{align} & \therefore r=\sqrt{{{(2-4)}^{2}}+{{(2-5)}^{2}}} \\ & =\sqrt{{{(-2)}^{2}}+{{(-3)}^{2}}} \\ & =\sqrt{4+9} \\ & =\sqrt{13} \\ \end{align}

Thus, the equation of the circle is

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

\begin{align} & {{(x-2)}^{2}}+{{(y-2)}^{2}}={{\left( \sqrt{13} \right)}^{2}} \\ & {{x}^{2}}-4x+4+{{y}^{2}}-4y+4=13 \\ & {{x}^{2}}+{{y}^{2}}-4x-4y-5=0 \\ \end{align}

15. Does the point $(-2.5,3.5)$ lie inside, outside or on the circle ${{x}^{2}}+{{y}^{2}}=25$?

Ans: The equation of the given circle is ${{x}^{2}}+{{y}^{2}}=25$.

${{x}^{2}}+{{y}^{2}}=25$

$\Rightarrow {{(x-0)}^{2}}+{{(y-0)}^{2}}={{5}^{2}},$ which is of the form of ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h = 0, k = 0 \; and\; r=5$

$\therefore$ Centre $=(0,0)$ and radius $=5$

Distance between point $(-2.5,3.5)$ and center $(0,0)$

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

\begin{align} & =\sqrt{6.25+12.25} \\ & =\sqrt{18.25} \\ \end{align}

$=4.3$ (approx.) $<5$

Since the distance between point $(-2.5,3.5)$ and center $(0,0)$ of the circle is less than the radius of the circle, point $(-2.5,3.5)$ lies inside the circle.

### Exercise 10.2

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

Ans: The given equation is ${{y}^{2}}=12x$.

Here, the coefficient of x is positive. Hence, the parabola opens towards the right. On comparing this equation with ${{y}^{2}}=4ax,$ we’ll get

$4a=12\Rightarrow a=3$

$\therefore$ Coordinates of the focus = $=(a,0)=(3,0)$

Since the given equation involves ${{y}^{2}}$, the axis of the parabola is the x-axis. Equation of directrix, $x=-a\text{ }i.e.,\text{ }x=-3\text{ }i.e.,\text{ }x+3=0$

Length of latus rectum $=4a=4\times 3=12$

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 ${{x}^{2}}=6y$.

Here, the coefficient of y is positive. Hence, the parabola opens upwards.

On comparing this equation with ${{x}^{2}}=4ay$ we obtain

$4a=6\Rightarrow a=\dfrac{3}{2}$

$\therefore$Coordinates of the focus $=(0,a)=\left( 0,\dfrac{3}{2} \right)$

Since the given equation involves ${{x}^{2}}$, the axis of the parabola is the y-axis. Equation of directrix, $y=-a$  i.e., $y=\dfrac{-3}{2}$

Length of latus rectum $=4a=6$

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

Ans: The given equation is ${{y}^{2}}=-8x$.

Here, the coefficient of $x$ is negative. Hence, the parabola opens towards the left. On comparing this equation with ${{y}^{2}}=-4ax,$ we’ll get

$-4a=-8\Rightarrow a=2$

$\therefore$ Coordinates of the focus $=(-a,0)=(-2,0)$

Since the given equation involves ${{y}^{2}}$, the axis of the parabola is the x-axis. Equation of directrix, $x=a$  i.e., $x=2$

Length of latus rectum $=4a=8$

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 ${{x}^{2}}=-16y.$

Here, the coefficient of $y$ is negative. Hence, the parabola opens downwards.

On comparing this equation with ${{x}^{2}}=-4ay$, we’ll get

$-4a=-16\Rightarrow a=4$

$\therefore$ Coordinates of the focus $=(0,-a)=(0,-4)$

Since the given equation involves ${{x}^{2}}$, the axis of the parabola is the y-axis. Equation of directrix, $y=a$  i.e., $y=4$

Length of latus rectum $=4a=16$

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

Ans:The given equation is ${{y}^{2}}=10x$.

Here, the coefficient of $x$ is positive. Hence, the parabola opens towards the right. On comparing this equation with ${{y}^{2}}=4ax$, we’ll get

$4a=10\Rightarrow a=\dfrac{5}{2}$

$\therefore$ Coordinates of the focus $=(a,0)=\left( \dfrac{5}{2},0 \right)$

Since the given equation involves ${{y}^{2}}$, the axis of the parabola is the x-axis. Equation of directrix, $x=-a$  i.e., $x=-\dfrac{5}{2}$

Length of latus rectum $=4a=10$

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 ${{x}^{2}}=-9y$.

Here, the coefficient of $y$ is negative. Hence, the parabola opens downwards.

On comparing this equation with ${{x}^{2}}=-4ay$, we’ll get

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

$\therefore$ Coordinates of the focus $=(0,-a)=\left( 0,-\dfrac{9}{4} \right)$

Since the given equation involves ${{x}^{2}}$, the axis of the parabola is the y-axis. Equation of directrix, $y=a$  i.e., $y=\dfrac{9}{4}$

Length of latus rectum $=4a=9$

7. Find the equation of the parabola that satisfies the following conditions: Focus $(6,0);$ directrix $x=-6$

Ans: Focus $(6,0);$directrix, $x=-6$

Since the focus lies on the x-axis, the x-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form ${{y}^{2}}=4ax$ or  ${{y}^{2}}=-4ax$.

It is also seen that the directrix, $x=-6$is to the left of the y-axis, while the focus $(6,0)$ is to the right of the y-axis.

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

Here, $a=6$

Thus, the equation of the parabola is ${{y}^{2}}=24x$.

8. Find the equation of the parabola that satisfies the following conditions: Focus $(0,-3);$ directrix $y=3$

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

Since the focus lies on the y-axis, the y-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form ${{x}^{2}}=4ay$ or ${{x}^{2}}=-4ay.$

It is also seen that the directrix, $y=3$ is above the x-axis, while the focus $(0,-3)$ is below the x-axis.

Hence, the parabola is of the form ${{x}^{2}}=-4ay.$

Here, $a=3$

Thus, the equation of the parabola is ${{x}^{2}}=-12y.$

9. Find the equation of the parabola that satisfies the following conditions: Vertex $(0,0);$ focus $(3,0)$

Ans: Vertex $(0,0);$ focus $(3,0)$

Since the vertex of the parabola is $(0,0)$ and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form ${{y}^{2}}=4ax.$

Since the focus is $(3,0)$, $a=3$.

Thus, the equation of the parabola is ${{y}^{2}}=4\times 3\times x$  i.e., ${{y}^{2}}=12x$

10. Find the equation of the parabola that satisfies the following conditions: Vertex $(0,0)$ focus $(-2,0)$

Ans: Solution 10: Vertex $(0,0)$ focus $(-2,0)$

Since the vertex of the parabola is $(0,0)$ and the focus lies on the negative x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form ${{y}^{2}}=-4ax.$

Since the focus is $(-2,0),$$a=2.$

Thus, the equation of the parabola is ${{y}^{2}}=-4\times 2\times x$  i.e., ${{y}^{2}}=-8x$

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

Ans: Since the vertex is (0, 0) and the axis of the parabola is the x-axis, the equation of the parabola is either of the form ${{y}^{2}}=4ax$ or ${{y}^{2}}=-4ax.$

The parabola passes through point $(2,3)$, which lies in the first quadrant. Therefore, the equation of the parabola is of the form ${{y}^{2}}=4ax$, while point $(2,3)$ must satisfy the equation ${{y}^{2}}=4ax$.

$\therefore {{(3)}^{2}}=4a(2)\Rightarrow a=\dfrac{9}{8}$

Thus, the equation of the parabola is ${{y}^{2}}=4\left( \dfrac{9}{8} \right)x$

\begin{align} & \Rightarrow {{y}^{2}}=\dfrac{9}{2}x \\ & \Rightarrow 2{{y}^{2}}=9x \\ \end{align}

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

Ans: Since the vertex is $(0,0)$ and the parabola is symmetric about the y-axis, the equation of the parabola is either of the form ${{x}^{2}}=4ay$ or ${{x}^{2}}=-4ay.$

The parabola passes through point $(5,2)$, which lies in the first quadrant. Therefore, the equation of the parabola is of the form ${{x}^{2}}=4ay$, while point $(5,2)$ must satisfy the equation ${{x}^{2}}=4ay$.

\begin{align} & \therefore {{(5)}^{2}}=4\times a\times 2 \\ & \Rightarrow 25=8a \\ & \Rightarrow a=\dfrac{25}{8} \\ \end{align}

Thus, the equation of the parabola is

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

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

### Exercise 10.3

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

Ans: The given equation is $\dfrac{{{x}^{2}}}{36}+\dfrac{{{y}^{2}}}{16}=1$

Here, the denominator of $\dfrac{{{x}^{2}}}{36}$ is greater than the denominator of $\dfrac{{{y}^{2}}}{16}$.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=6$ and $b=4$

\begin{align} & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\ & =\sqrt{36-16} \\ & =\sqrt{20}=2\sqrt{5} \\ \end{align}

Therefore,

The coordinates of the foci are $\left( 2\sqrt{5},0 \right)$ and $\left( -2\sqrt{5},0 \right)$

The coordinates of the vertices are $\left( 6,0 \right)$and$\left( -6,0 \right)$.

Length of major axis $=2a=12$

Length of minor axis $=2b=8$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{2\sqrt{5}}{6}=\dfrac{\sqrt{5}}{3}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 16}{6}=\dfrac{16}{3}$

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

Ans: The given equation is $\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{25}=1$ or $\dfrac{{{x}^{2}}}{{{2}^{2}}}+\dfrac{{{y}^{2}}}{{{5}^{2}}}=1$

Here, the denominator of $\dfrac{{{y}^{2}}}{25}$ is greater than the denominator of $\dfrac{{{x}^{2}}}{4}$.

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with $\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1$, we’ll get $b=2$ and $a=5$

\begin{align} & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\ & =\sqrt{25-4} \\ & =\sqrt{21} \\ \end{align}

Therefore,

The coordinates of the foci are $\left( 0,\sqrt{21} \right)$ and $\left( 0,-\sqrt{21} \right)$

The coordinates of the vertices are $(0,5)$ and $(0,-5)$

Length of major axis $=2a=10$

Length of minor axis $=2b=4$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{21}}{5}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 4}{5}=\dfrac{8}{5}$

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

Ans: The given equation is $\dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{9}=1$ or  $\dfrac{{{x}^{2}}}{{{4}^{2}}}+\dfrac{{{y}^{2}}}{{{3}^{2}}}=1$

Here, the denominator of $\dfrac{{{x}^{2}}}{16}$ is greater than the denominator of $\dfrac{{{y}^{2}}}{9}$.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=4$ and $b=3$

\begin{align} & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\ & =\sqrt{16-9} \\ & =\sqrt{7} \\ \end{align}

Therefore,

The coordinates of the foci are $\left( \pm \sqrt{7},0 \right)$

The coordinates of the vertices are $(\pm 4,0)$

Length of major axis $=2a=8$

Length of minor axis $=2b=6$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{7}}{4}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 9}{4}=\dfrac{9}{2}$

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

Ans: The given equation is $\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{100}=1$ or  $\dfrac{{{x}^{2}}}{{{5}^{2}}}+\dfrac{{{y}^{2}}}{{{10}^{2}}}=1$

Here, the denominator of $\dfrac{{{y}^{2}}}{100}$ is greater than the denominator of $\dfrac{{{x}^{2}}}{25}$.

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with $\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1$, we’ll get $b=5$ and $a=10$

\begin{align} & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\ & =\sqrt{100-25} \\ & =\sqrt{75} \\ & =5\sqrt{3} \\ \end{align}

Therefore,

The coordinates of the foci are $\left( 0,\pm 5\sqrt{3} \right)$

The coordinates of the vertices are $(0,\pm 10)$

Length of major axis $=2a=20$

Length of minor axis $=2b=10$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{5\sqrt{3}}{10}=\dfrac{\sqrt{3}}{2}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 25}{10}=5$

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

Ans: The given equation is $\dfrac{{{x}^{2}}}{49}+\dfrac{{{y}^{2}}}{36}=1$ or  $\dfrac{{{x}^{2}}}{{{7}^{2}}}+\dfrac{{{y}^{2}}}{{{6}^{2}}}=1$

Here, the denominator of $\dfrac{{{x}^{2}}}{49}$ is greater than the denominator of $\dfrac{{{y}^{2}}}{36}$.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=7$and $b=6$

\begin{align} & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\ & =\sqrt{49-36} \\ & =\sqrt{13} \\ \end{align}

Therefore,

The coordinates of the foci are $\left( \pm \sqrt{13},0 \right)$

The coordinates of the vertices are $(\pm 7,0)$

Length of major axis $=2a=14$

Length of minor axis $=2b=12$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{13}}{7}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 36}{7}=\dfrac{72}{7}$

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

Ans: The given equation is $\dfrac{{{x}^{2}}}{100}+\dfrac{{{y}^{2}}}{400}=1$ or  $\dfrac{{{x}^{2}}}{{{10}^{2}}}+\dfrac{{{y}^{2}}}{{{20}^{2}}}=1$

Here, the denominator of $\dfrac{{{y}^{2}}}{400}$ is greater than the denominator of $\dfrac{{{x}^{2}}}{100}$.

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with $\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1$, we’ll get $b=10$ and $a=20$

\begin{align} & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\ & =\sqrt{400-100} \\ & =\sqrt{300} \\ & =10\sqrt{3} \\ \end{align}

Therefore,

The coordinates of the foci are $\left( 0,\pm 10\sqrt{3} \right)$

The coordinates of the vertices are $(0,\pm 20)$

Length of major axis $=2a=40$

Length of minor axis $=2b=20$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{10\sqrt{3}}{20}=\dfrac{\sqrt{3}}{2}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 100}{20}=10$

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

Ans: The given equation is $36{{x}^{2}}+4{{y}^{2}}=144$.

It can be written as

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

Or, $\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{36}=1$

Or, $\dfrac{{{x}^{2}}}{{{2}^{2}}}+\dfrac{{{y}^{2}}}{{{6}^{2}}}=1$ ………(1)

Here, the denominator of $\dfrac{{{y}^{2}}}{{{6}^{2}}}$is greater than the denominator of $\dfrac{{{x}^{2}}}{{{2}^{2}}}$.

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with $\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1$, we’ll get $b=2$ and $a=6$

\begin{align} & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\ & =\sqrt{36-4} \\ & =\sqrt{32} \\ & =4\sqrt{2} \\ \end{align}

Therefore,

The coordinates of the foci are $\left( 0,\pm 4\sqrt{2} \right)$

The coordinates of the vertices are $(0,\pm 6)$

Length of major axis $=2a=12$

Length of minor axis $=2b=4$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{4\sqrt{2}}{6}=\dfrac{2\sqrt{2}}{3}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 4}{6}=\dfrac{4}{3}$

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

Ans: The given equation is $16{{x}^{2}}+{{y}^{2}}=16$.

It can be written as

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

Or, $\dfrac{{{x}^{2}}}{1}+\dfrac{{{y}^{2}}}{16}=1$

Or, $\dfrac{{{x}^{2}}}{{{1}^{2}}}+\dfrac{{{y}^{2}}}{{{4}^{2}}}=1$ ………(1)

Here, the denominator of $\dfrac{{{y}^{2}}}{{{4}^{2}}}$is greater than the denominator of $\dfrac{{{x}^{2}}}{{{1}^{2}}}$.

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with $\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1$, we’ll get $b=1$ and $a=4$

\begin{align} & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\ & =\sqrt{16-1} \\ & =\sqrt{15} \\ \end{align}

Therefore,

The coordinates of the foci are $\left( 0,\pm \sqrt{15} \right)$

The coordinates of the vertices are $(0,\pm 4)$

Length of major axis $=2a=8$

Length of minor axis $=2b=2$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{15}}{6}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 1}{4}=\dfrac{1}{2}$

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

Ans: The given equation is $4{{x}^{2}}+9{{y}^{2}}=36$.

It can be written as

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

Or, $\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}=1$

Or, $\dfrac{{{x}^{2}}}{{{3}^{2}}}+\dfrac{{{y}^{2}}}{{{2}^{2}}}=1$ ………(1)

Here, the denominator of $\dfrac{{{x}^{2}}}{{{3}^{2}}}$is greater than the denominator of $\dfrac{{{y}^{2}}}{{{2}^{2}}}$.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=3$ and $b=2$

\begin{align} & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\ & =\sqrt{9-4} \\ & =\sqrt{5} \\ \end{align}

Therefore,

The coordinates of the foci are $\left( \pm \sqrt{5},0 \right)$

The coordinates of the vertices are $(\pm 3,0)$

Length of major axis $=2a=6$

Length of minor axis $=2b=4$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{5}}{3}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 4}{3}=\dfrac{8}{3}$

10. Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 5,0),$foci $(\pm 4,0)$.

Ans: Vertices $(\pm 5,0),$ foci $(\pm 4,0)$

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, where a is the semi-major axis.

Accordingly, $a=5$ and $c=4$.

It is known that ${{a}^{2}}={{b}^{2}}+{{c}^{2}}$.

\begin{align} & \therefore {{5}^{2}}={{b}^{2}}+{{4}^{2}} \\ & \Rightarrow 25={{b}^{2}}+16 \\ & \Rightarrow b=\sqrt{9}=3 \\ \end{align}

Thus, the equation of the ellipse is $\dfrac{{{x}^{2}}}{{{5}^{2}}}+\dfrac{{{y}^{2}}}{{{3}^{2}}}=1$ or $\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{9}=1$

11. Find the equation for the ellipse that satisfies the given conditions: Vertices $(0,\pm 13),$ foci $(0,\pm 5)$

Ans:  Vertices $(0,\pm 13),$ foci $(0,\pm 5)$

Here, the vertices are on the y-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1$, where a is the semi-major axis.

Accordingly, $a=13$ and $c=5$.

It is known that ${{a}^{2}}={{b}^{2}}+{{c}^{2}}$.

\begin{align} & \therefore {{13}^{2}}={{b}^{2}}+{{5}^{2}} \\ & \Rightarrow 169={{b}^{2}}+25 \\ & \Rightarrow b=\sqrt{144}=12 \\ \end{align}

Thus, the equation of the ellipse is $\dfrac{{{x}^{2}}}{{{12}^{2}}}+\dfrac{{{y}^{2}}}{{{13}^{2}}}=1$ or $\dfrac{{{x}^{2}}}{144}+\dfrac{{{y}^{2}}}{169}=1$

12. Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 6,0),$ foci $(\pm 4,0)$

Ans: Vertices $(\pm 6,0),$ foci $(\pm 4,0)$

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, where a is the semi-major axis.

Accordingly, $a=6$ and $c=4$.

It is known that ${{a}^{2}}={{b}^{2}}+{{c}^{2}}$.

\begin{align} & \therefore {{6}^{2}}={{b}^{2}}+{{4}^{2}} \\ & \Rightarrow 36={{b}^{2}}+16 \\ & \Rightarrow b=\sqrt{20} \\ \end{align}

Thus, the equation of the ellipse is $\dfrac{{{x}^{2}}}{{{6}^{2}}}+\dfrac{{{y}^{2}}}{{{\left( \sqrt{20} \right)}^{2}}}=1$ or $\dfrac{{{x}^{2}}}{36}+\dfrac{{{y}^{2}}}{20}=1$

13. Find the equation for the ellipse that satisfies the given conditions: Ends of major axis $(\pm 3,0),$ends of minor axis $(0,\pm 2)$

Ans: Ends of major axis $(\pm 3,0),$ ends of minor axis $(0,\pm 2)$

Here, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, where a is the semimajor axis.

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

Thus, the equation of the ellipse is $\dfrac{{{x}^{2}}}{{{3}^{2}}}+\dfrac{{{y}^{2}}}{{{2}^{2}}}=1$or $\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}=1$

14. Find the equation for the ellipse that satisfies the given conditions: Ends of major axis $(0,\pm \sqrt{5}),$ ends of minor axis $(\pm 1,0)$

Ans: Ends of major axis $(0,\pm \sqrt{5}),$ ends of minor axis $(\pm 1,0)$

Here, the major axis is along the y-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1$, where a is the semimajor axis.

Accordingly, $a=\sqrt{5}$ and $b=1$.

Thus, the equation of the ellipse is $\dfrac{{{x}^{2}}}{{{1}^{2}}}+\dfrac{{{y}^{2}}}{{{\left( \sqrt{5} \right)}^{2}}}=1$ or $\dfrac{{{x}^{2}}}{1}+\dfrac{{{y}^{2}}}{5}=1$

15. Find the equation for the ellipse that satisfies the given conditions: Length of major axis $26$, foci $(\pm 5,0)$

Ans: Length of major axis = $26$; foci = $(\pm 5,0)$

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, where a is the semimajor axis.

Accordingly, $2a=26\Rightarrow a=13$ and $c=5$.

It is known that ${{a}^{2}}={{b}^{2}}+{{c}^{2}}$.

\begin{align} & \therefore {{13}^{2}}={{b}^{2}}+{{5}^{2}} \\ & \Rightarrow 169={{b}^{2}}+25 \\ & \Rightarrow b=\sqrt{144}=12 \\ \end{align}

Thus, the equation of the ellipse is $\dfrac{{{x}^{2}}}{{{13}^{2}}}+\dfrac{{{y}^{2}}}{{{12}^{2}}}=1$or $\dfrac{{{x}^{2}}}{169}+\dfrac{{{y}^{2}}}{144}=1$

16. Find the equation for the ellipse that satisfies the given conditions: Length of minor axis $16$, foci $(0,\pm 6)$

Ans: Length of minor axis = $16$, foci = $(0,\pm 6)$

Since the foci are on the y-axis, the major axis is along the y-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1$, where a is the semimajor axis.

Accordingly, $2b=16\Rightarrow b=8$ and $c=6$.

It is known that ${{a}^{2}}={{b}^{2}}+{{c}^{2}}$.

\begin{align} & \therefore {{a}^{2}}={{8}^{2}}+{{6}^{2}} \\ & \Rightarrow {{a}^{2}}=64+36 \\ & \Rightarrow a=\sqrt{100}=10 \\ \end{align}

Thus, the equation of the ellipse is $\dfrac{{{x}^{2}}}{{{8}^{2}}}+\dfrac{{{y}^{2}}}{{{10}^{2}}}=1$ or $\dfrac{{{x}^{2}}}{64}+\dfrac{{{y}^{2}}}{100}=1$

17. Find the equation for the ellipse that satisfies the given conditions: Foci $(\pm 3,0),$ $a=4$

Ans: Foci $(\pm 3,0),$$a=4$

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, where a is the semimajor axis.

Accordingly, $c=3$ and $a=4$.

It is known that ${{a}^{2}}={{b}^{2}}+{{c}^{2}}$.

\begin{align} & \therefore {{4}^{2}}={{b}^{2}}+{{3}^{2}} \\ & \Rightarrow 16={{b}^{2}}+9 \\ & \Rightarrow {{b}^{2}}=16-9=7 \\ \end{align}

Thus, the equation of the ellipse is $\dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{7}=1$

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.

Ans: It is given that $b=3$, $c=4$, centre at the origin; foci on the x axis.

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, where a is the semimajor axis.

Accordingly, $b=3$, $c=4$.

It is known that ${{a}^{2}}={{b}^{2}}+{{c}^{2}}$.

\begin{align} & \therefore {{a}^{2}}={{3}^{2}}+{{4}^{2}} \\ & \Rightarrow {{a}^{2}}=9+16 \\ & \Rightarrow {{a}^{2}}=25 \\ & \Rightarrow a=\sqrt{25}=5 \\ \end{align}

Thus, the equation of the ellipse is $\dfrac{{{x}^{2}}}{{{5}^{2}}}+\dfrac{{{y}^{2}}}{{{3}^{2}}}=1$ or $\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{9}=1$

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 $(1,6)$.

Ans: Since the centre is at $(0,0)$ and the major axis is on the y-axis, the equation of the ellipse will be of the form

$\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1$ …..(1)

Where, a is the semi-major axis

The ellipse passes through points $(3,2)$ and $(1,6)$. Hence,

$\dfrac{9}{{{b}^{2}}}+\dfrac{4}{{{a}^{2}}}=1$ …..(2)

$\dfrac{1}{{{b}^{2}}}+\dfrac{36}{{{a}^{2}}}=1$ …..(3)

On solving equations (2) and (3), we’ll get

${{b}^{2}}=10$ and ${{a}^{2}}=40$.

Thus, the equation of the ellipse is $\dfrac{{{x}^{2}}}{10}+\dfrac{{{y}^{2}}}{40}=1$ or $4{{x}^{2}}+{{y}^{2}}=40$.

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 $(6,2)$.

Ans: Since the major axis is on the x-axis, the equation of the ellipse will be of the form

$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ …..(1)

Where, a is the semi-major axis

The ellipse passes through points $(4,3)$ and $(6,2)$. Hence,

$\dfrac{16}{{{a}^{2}}}+\dfrac{9}{{{b}^{2}}}=1$ …..(2)

$\dfrac{36}{{{a}^{2}}}+\dfrac{4}{{{b}^{2}}}=1$ …..(3)

On solving equations (2) and (3), we’ll get

${{a}^{2}}=52$ and ${{b}^{2}}=13$.

Thus, the equation of the ellipse is $\dfrac{{{x}^{2}}}{52}+\dfrac{{{y}^{2}}}{13}=1$ or ${{x}^{2}}+4{{y}^{2}}=52$.

### Exercise 10.4

1. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{9}=1$

Ans: The given equation is $\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{9}=1$ or $\dfrac{{{x}^{2}}}{{{4}^{2}}}-\dfrac{{{y}^{2}}}{{{3}^{2}}}=1$

On comparing this equation with the standard equation of hyperbola i.e., $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=4$ and $b=3$.

We know that ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$.

\begin{align} & \therefore {{c}^{2}}={{4}^{2}}+{{3}^{2}} \\ & \Rightarrow {{c}^{2}}=16+9 \\ & \Rightarrow {{c}^{2}}=25 \\ & \Rightarrow c=\sqrt{25}=5 \\ \end{align}

Therefore,

The coordinates of the foci are $\left( \pm 5,0 \right)$

The coordinates of the vertices are $(\pm 4,0)$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{5}{4}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 9}{4}=\dfrac{9}{2}$

2. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\dfrac{{{y}^{2}}}{9}-\dfrac{{{x}^{2}}}{27}=1$

Ans: The given equation is $\dfrac{{{y}^{2}}}{9}-\dfrac{{{x}^{2}}}{27}=1$ or $\dfrac{{{y}^{2}}}{{{3}^{2}}}-\dfrac{{{x}^{2}}}{{{\left( \sqrt{27} \right)}^{2}}}=1$

On comparing this equation with the standard equation of hyperbola i.e., $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=3$ and $b=\sqrt{27}$.

We know that ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$.

\begin{align} & \therefore {{c}^{2}}={{3}^{2}}+{{\left( \sqrt{27} \right)}^{2}} \\ & \Rightarrow {{c}^{2}}=9+27 \\ & \Rightarrow {{c}^{2}}=36 \\ & \Rightarrow c=\sqrt{36}=6 \\ \end{align}

Therefore,

The coordinates of the foci are $\left( 0,\pm 6 \right)$

The coordinates of the vertices are $(0,\pm 3)$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{6}{3}=2$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 27}{3}=18$

3. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $9{{y}^{2}}-4{{x}^{2}}=36$

Ans: The given equation is $9{{y}^{2}}-4{{x}^{2}}=36$.

It can be written as

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

Or, $\dfrac{{{y}^{2}}}{4}-\dfrac{{{x}^{2}}}{9}=1$

Or, $\dfrac{{{y}^{2}}}{{{2}^{2}}}-\dfrac{{{x}^{2}}}{{{3}^{2}}}=1$ ………(1)

On comparing equation (1) with the standard equation of hyperbola i.e., $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=2$ and $b=3$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

\begin{align} & \therefore {{c}^{2}}=4+9 \\ & \Rightarrow {{c}^{2}}=13 \\ & \Rightarrow c=\sqrt{13} \\ \end{align}

Therefore,

The coordinates of the foci are $\left( 0,\pm \sqrt{13} \right)$

The coordinates of the vertices are $(0,\pm 2)$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{13}}{2}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 9}{2}=9$

4. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $16{{x}^{2}}-9{{y}^{2}}=576$

Ans: The given equation is $16{{x}^{2}}-9{{y}^{2}}=576$.

It can be written as

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

Or, $\dfrac{{{x}^{2}}}{36}-\dfrac{{{y}^{2}}}{64}=1$

Or, $\dfrac{{{x}^{2}}}{{{6}^{2}}}-\dfrac{{{y}^{2}}}{{{8}^{2}}}=1$ ………(1)

On comparing equation (1) with the standard equation of hyperbola i.e., $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=6$ and $b=8$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

\begin{align} & \therefore {{c}^{2}}=36+64 \\ & \Rightarrow {{c}^{2}}=100 \\ & \Rightarrow c=\sqrt{100} \\ & \Rightarrow c=10 \\ \end{align}

Therefore,

The coordinates of the foci are $\left( \pm \sqrt{10},0 \right)$

The coordinates of the vertices are $(\pm 6,0)$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{10}{6}=\dfrac{5}{3}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 64}{6}=\dfrac{64}{3}$

5. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $5{{y}^{2}}-9{{x}^{2}}=36$

Ans: The given equation is $5{{y}^{2}}-9{{x}^{2}}=36$.

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

$\dfrac{{{y}^{2}}}{\left( \dfrac{6}{\sqrt{5}} \right)^2}-\dfrac{{{x}^{2}}}{{{2}^{2}}}=1$ ………(1)

On comparing equation (1) with the standard equation of hyperbola i.e., $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=\dfrac{6}{\sqrt{5}}$ and $b=2$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

\begin{align} & \therefore {{c}^{2}}=\dfrac{36}{5}+4 \\ & \Rightarrow {{c}^{2}}=\dfrac{56}{5} \\ & \Rightarrow c=\sqrt{\dfrac{56}{5}} \\ & \Rightarrow c=\dfrac{2\sqrt{14}}{\sqrt{5}} \\ \end{align}

Therefore,

The coordinates of the foci are $\left( 0,\pm \dfrac{2\sqrt{14}}{\sqrt{5}} \right)$

The coordinates of the vertices are $\left( 0,\pm \dfrac{6}{\sqrt{5}} \right)$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\left( \dfrac{2\sqrt{14}}{\sqrt{5}} \right)}{\left( \dfrac{6}{\sqrt{5}} \right)}=\dfrac{\sqrt{14}}{3}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 4}{\left( \dfrac{6}{\sqrt{5}} \right)}=\dfrac{4\sqrt{5}}{3}$

6. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $49{{y}^{2}}-16{{x}^{2}}=784$

Ans: The given equation is $49{{y}^{2}}-16{{x}^{2}}=784$.

It can be written as

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

Or, $\dfrac{{{y}^{2}}}{16}-\dfrac{{{x}^{2}}}{49}=1$

Or, $\dfrac{{{y}^{2}}}{{{4}^{2}}}-\dfrac{{{x}^{2}}}{{{7}^{2}}}=1$ ………(1)

On comparing equation (1) with the standard equation of hyperbola i.e., $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=4$ and $b=7$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

\begin{align} & \therefore {{c}^{2}}=16+49 \\ & \Rightarrow {{c}^{2}}=65 \\ & \Rightarrow c=\sqrt{65} \\ \end{align}

Therefore,

The coordinates of the foci are $\left( 0,\pm \sqrt{65} \right)$

The coordinates of the vertices are $(0,\pm 4)$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{65}}{4}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 49}{4}=\dfrac{49}{2}$

7. Find the equation of the hyperbola satisfying the give conditions: Vertices $(\pm 2,0)$, foci $(\pm 3,0)$

Ans: Vertices $(\pm 2,0)$, foci $(\pm 3,0)$

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$

Since the vertices are $(\pm 2,0)$, $a=2$.

Since the foci are $(\pm 3,0)$, $c=3$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

\begin{align} & \therefore {{2}^{2}}+{{b}^{2}}={{3}^{2}} \\ & \Rightarrow {{b}^{2}}=9-4 \\ & \Rightarrow {{b}^{2}}=5 \\ \end{align}

Thus, the equation of the hyperbola is $\dfrac{{{x}^{2}}}{4}-\dfrac{{{y}^{2}}}{5}=1$

8. Find the equation of the hyperbola satisfying the give conditions: Vertices $(0,\pm 5)$, foci $(0,\pm 8)$

Ans: Vertices $(0,\pm 5)$, foci $(0,\pm 8)$

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$

Since the vertices are $(0,\pm 5)$, $a=5$.

Since the foci are $(0,\pm 8)$, $c=8$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

\begin{align} & \therefore {{5}^{2}}+{{b}^{2}}={{8}^{2}} \\ & \Rightarrow {{b}^{2}}=64-25 \\ & \Rightarrow {{b}^{2}}=39 \\ \end{align}

Thus, the equation of the hyperbola is $\dfrac{{{y}^{2}}}{25}-\dfrac{{{x}^{2}}}{39}=1$

9. Find the equation of the hyperbola satisfying the give conditions: Vertices $(0,\pm 3)$, foci $(0,\pm 5)$

Ans: Vertices $(0,\pm 3)$, foci $(0,\pm 5)$

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$

Since the vertices are $(0,\pm 3)$, $a=3$.

Since the foci are $(0,\pm 5)$, $c=5$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

\begin{align} & \therefore {{3}^{2}}+{{b}^{2}}=25 \\ & \Rightarrow {{b}^{2}}=25-9 \\ & \Rightarrow {{b}^{2}}=16 \\ \end{align}

Thus, the equation of the hyperbola is $\dfrac{{{y}^{2}}}{9}-\dfrac{{{x}^{2}}}{16}=1$

10. Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 5,0)$, the transverse axis is of length $8$.

Ans: Foci $(\pm 5,0)$, the transverse axis is of length $8$.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.

Since the foci are $(\pm 5,0)$, $c=5$.

Since the length of the transverse axis is $8$, $2a=8\Rightarrow a=4$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

\begin{align} & \therefore {{4}^{2}}+{{b}^{2}}=25 \\ & \Rightarrow {{b}^{2}}=25-16 \\ & \Rightarrow {{b}^{2}}=9 \\ \end{align}

Thus, the equation of the hyperbola is $\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{9}=1$

11. Find the equation of the hyperbola satisfying the give conditions: Foci $(0,\pm 13)$, the conjugate axis is of length $24$.

Ans: Foci $(0,\pm 13)$, the transverse axis is of length $24$.

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$.

Since the foci are $(0,\pm 13)$, $c=13$.

Since the length of the transverse axis is $24$, $2b=24\Rightarrow b=12$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

\begin{align} & \therefore {{a}^{2}}+{{12}^{2}}={{13}^{2}} \\ & \Rightarrow {{a}^{2}}=169-144 \\ & \Rightarrow {{a}^{2}}=25 \\ \end{align}

Thus, the equation of the hyperbola is $\dfrac{{{y}^{2}}}{25}-\dfrac{{{x}^{2}}}{144}=1$

12. Find the equation of the hyperbola satisfying the give conditions: Foci $\left( \pm 3\sqrt{5},0 \right)$, the latus rectum is of length $8$.

Ans: Foci $\left( \pm 3\sqrt{5},0 \right)$, the latus rectum is of length $8$.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$

Since the foci are $\left( \pm 3\sqrt{5},0 \right)$, $c=\pm 3\sqrt{5}$

Length of latus rectum = $8$

\begin{align} & \Rightarrow \dfrac{2{{b}^{2}}}{a}=8 \\ & \Rightarrow {{b}^{2}}=4a \\ \end{align}

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

\begin{align} & \therefore {{a}^{2}}+4a=45 \\ & \Rightarrow {{a}^{2}}+4a-45=0 \\ & \Rightarrow {{a}^{2}}+9a-5a-45=0 \\ & \Rightarrow \left( a+9 \right)\left( a-5 \right)=0 \\ & \Rightarrow a=-9,5 \\ \end{align}

Since a is non-negative, $a=5$.

$\therefore {{b}^{2}}=4a=4\times 5=20$

Thus, the equation of the hyperbola is $\dfrac{{{x}^{2}}}{25}-\dfrac{{{y}^{2}}}{20}=1$

13. Find the equation of the hyperbola satisfying the give conditions: Foci $\left( \pm 4,0 \right)$, the latus rectum is of length $12$.

Ans: Foci $\left( \pm 4,0 \right)$, the latus rectum is of length $12$.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$

Since the foci are $\left( \pm 4,0 \right)$, $c=4$.

Length of latus rectum = $12$

\begin{align} & \Rightarrow \dfrac{2{{b}^{2}}}{a}=12 \\ & \Rightarrow {{b}^{2}}=6a \\ \end{align}

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

\begin{align} & \therefore {{a}^{2}}+6a=16 \\ & \Rightarrow {{a}^{2}}+6a-16=0 \\ & \Rightarrow {{a}^{2}}+8a-2a-16=0 \\ & \Rightarrow \left( a+8 \right)\left( a-2 \right)=0 \\ & \Rightarrow a=-8,2 \\ \end{align}

Since a is non-negative, $a=2$.

$\therefore {{b}^{2}}=6a=6\times 2=12$

Thus, the equation of the hyperbola is $\dfrac{{{x}^{2}}}{4}-\dfrac{{{y}^{2}}}{12}=1$

14. Find the equation of the hyperbola satisfying the give conditions: Vertices $\left( \pm 7,0 \right)$, $e=\dfrac{4}{3}$

Ans: Vertices $\left( \pm 7,0 \right)$, $e=\dfrac{4}{3}$

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$

Since the vertices are $\left( \pm 7,0 \right)$, $a=7$.

It is given that $e=\dfrac{4}{3}$

$\therefore \dfrac{c}{a}=\dfrac{4}{3}$ $\left[ e=\dfrac{c}{a} \right]$

\begin{align} & \Rightarrow \dfrac{c}{7}=\dfrac{4}{3} \\ & \Rightarrow c=\dfrac{28}{3} \\ \end{align}

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

\begin{align} & \therefore {{7}^{2}}+{{b}^{2}}={{\left( \dfrac{28}{3} \right)}^{2}} \\ & \Rightarrow {{b}^{2}}=\dfrac{784}{9}-49 \\ & \Rightarrow {{b}^{2}}=\dfrac{784-441}{9}=\dfrac{343}{9} \\ \end{align}

Thus, the equation of the hyperbola is $\dfrac{{{x}^{2}}}{49}-\dfrac{{9{y}^{2}}}{343}=1$

15. Find the equation of the hyperbola satisfying the given conditions: Foci $\left( 0,\pm \sqrt{10} \right)$, passing through $\left( 2,3 \right)$

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

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$ .

Since the foci are $\left( 0,\pm \sqrt{10} \right)$, $c=\sqrt{10}$ .

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

$\therefore {{a}^{2}}+{{b}^{2}}=10$

${{a}^{2}}+{{b}^{2}}={{c}^{2}}$.

$\Rightarrow {{b}^{2}}=10-{{a}^{2}}$ … (1)

Since the hyperbola passes through point $\left( 2,3 \right)$,

$\dfrac{9}{{{a}^{2}}}-\dfrac{4}{{{b}^{2}}}=1$ ….(2)

From equations (1) and (2), we’ll get

\begin{align} & \dfrac{9}{{{a}^{2}}}-\dfrac{4}{{{\left( 10-a^{2}\right)}}}=1 \\ & \Rightarrow 9\left( 10-{{a}^{2}} \right)-4{{a}^{2}}={{a}^{2}}\left( 10-{{a}^{2}} \right) \\ & \Rightarrow 90-9{{a}^{2}}-4{{a}^{2}}=10{{a}^{2}}-{{a}^{4}} \\ & \Rightarrow {{a}^{4}}-10{{a}^{2}}-9{{a}^{2}}-4{{a}^{2}}+90=0 \\ & \Rightarrow {{a}^{4}}-23{{a}^{2}}+90=0 \\ & \Rightarrow {{a}^{4}}-18{{a}^{2}}-5{{a}^{2}}+90=0 \\ & \Rightarrow {{a}^{2}}({{a}^{2}}-18)-5({{a}^{2}}-18)=0 \\ & \Rightarrow ({{a}^{2}}-5)({{a}^{2}}-18)=0 \\ & \Rightarrow {{a}^{2}}=18,5 \\ \end{align}

In hyperbola, $c>a$ i.e., ${{c}^{2}}>{{a}^{2}}$

\begin{align} & \therefore a{}^{2}=5 \\ & \Rightarrow {{b}^{2}}=10-{{a}^{2}} \\ & \Rightarrow {{b}^{2}}=10-5=5 \\ \end{align}

Thus, the equation of the hyperbola is $\dfrac{{{y}^{2}}}{5}-\dfrac{{{x}^{2}}}{5}=1$

### Miscellaneous Exercise

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

Ans: As we know that the origin of the coordinate plane is taken at the vertex of the parabolic reflector, where the axis of the reflector is along the positive x-axis.

The Diagrammatic representation is represented as follows:

As we know that the equation of the parabola is of the form of ${{y}^{2}}=4ax$ (as it is opening to the right)

Since, the parabola passes through point $A \left( 5,10 \right),$

\begin{align} & \Rightarrow {{y}^{2}}=4ax \\ & \Rightarrow {{10}^{2}}=4\times a\times 5 \\ & \Rightarrow 100=20a \\ & \Rightarrow a=\dfrac{100}{20} \\ & \Rightarrow a=5 \\ \end{align}

The focus of the parabola is $\left( a,0 \right)=\left( 5,0 \right),$ which is the mid–point of the diameter.

Hence, the focus of the reflector is at the midpoint of the diameter.

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?

Ans: As we know that the origin of the coordinate plane is taken at the vertex of the arch in such a way that its axis is along the y-axis.

The diagrammatic representation will be as follows:

The equation of the parabola is of the form ${{x}^{2}}=-4ay$ (as it is opening downwards).

Since the parabola passes through point $\left( \dfrac{5}{2},-10 \right),$

\begin{align} & {{\left( \dfrac{5}{2} \right)}^{2}}=-4\times a\times (-10) \\ & \Rightarrow a=\dfrac{25}{4\times 4\times 10}=\dfrac{5}{32} \\ \end{align}

Therefore,

The arch is in the form of a parabola whose equation is ${{x}^{2}}=-\dfrac{5}{8}y$

When $y=-2m,{{x}^{2}}=-\dfrac{5}{8}\times (-2)$ because it's 2m away downward from the vertex, so it will be negative.

\begin{align} & \Rightarrow {{x}^{2}}=\dfrac{5}{4} \\ & \Rightarrow x=\sqrt{\dfrac{5}{4}}m \\ \end{align}

Width of parabola at 2m away from vertex will be $=2\times \sqrt{\dfrac{5}{4}}m$

$=2\times 1.118m$(approx.)

$=2.23m$(approx.)

Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately 2.23m.

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.

Ans: The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y-axis.

This can be diagrammatically represented as

Here, AB and OC are the longest and shortest wires, respectively, attached to the cable.

DF is the supporting wire attached to the roadway, 18m from the middle.

Here, AB$=30m, OC=6m, and\; BC=\dfrac{100}{2}=50m$

The equation of the parabola is of the form ${{x}^{2}}=4ay$ (as it is opening upwards).

The coordinates of point A are $(50,30-6)=(50,24)$.

Since A$(50,24)$ is a point on the parabola,

\begin{align} & {{(50)}^{2}}=4a(24) \\ & \Rightarrow a=\dfrac{50\times 50}{4\times 24}=\dfrac{625}{24} \\ \end{align}

$\therefore$Equation of the parabola, ${{x}^{2}}=4\times \dfrac{625}{24}\times y$ or $6{{x}^{2}}=625y$

The x-coordinate of point D is 18.

Hence, at x = 18,

\begin{align} & 6{{(18)}^{2}}=625y \\ & \Rightarrow y=\dfrac{6\times 18\times 18}{625} \\ & \Rightarrow y=3.11(approx.) \\ \end{align}

$\therefore$DE = $3.11$m

DF = DE + EF = $3.11$ m + $6$ m = $9.11$ m

Thus, the length of the supporting wire attached to the roadway 18 m from the middle is approximately $9.11$ m.

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.

Ans: Since the height and width of the arc from the centre is 2 m and 8 m respectively, it is clear that the length of the major axis is 8 m, while the length of the semi-minor axis is 2 m.

The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis.

Hence, the semi-ellipse can be diagrammatically represented as

The equation of the semi-ellipse will be of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1,y\ge 0,$ where a is the semi major axis

Accordingly,

\begin{align} & 2a=8\Rightarrow a=4 \\ & b=2 \\ \end{align}

Therefore, the equation of the semi-ellipse is $\dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{4}=1,y\ge 0,$ ….(1)

Let A be a point on the major axis such that AB = 1.5 m.

Draw AC ⊥ OB.

OA = (4 – 1.5) m = 2.5 m

The x-coordinate of point C is 2.5.

On substituting the value of x with 2.5 in equation (1), we’ll get

\begin{align} & \dfrac{{{\left( 2.5 \right)}^{2}}}{16}+\dfrac{{{y}^{2}}}{4}=1 \\ & \Rightarrow \dfrac{6.25}{16}+\dfrac{{{y}^{2}}}{4}=1 \\ & \Rightarrow {{y}^{2}}=4\left( 1-\dfrac{6.25}{16} \right) \\ & \Rightarrow {{y}^{2}}=4\left( \dfrac{9.75}{16} \right) \\ & \Rightarrow {{y}^{2}}=2.4375 \\ & \Rightarrow {y}=1.56(approx.) \\ & \therefore AC=1.56m \\ \end{align}

Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56 m.

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.

Ans: Let AB be the rod making an angle θ with OX and P (x, y) be the point on it such that AP = 3 cm.

Then, PB = AB – AP = (12 – 3) cm = 9 cm [AB = 12 cm]

From P, draw PQ⊥OY and PR⊥OX.

In ΔPBQ, $\cos \theta =\dfrac{PQ}{PB}=\dfrac{x}{9}$

In ΔPRA, $\sin \theta =\dfrac{PR}{PA}=\dfrac{y}{3}$

Since, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$

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

Or,  ${{\dfrac{x}{81}}^{2}}+{{\dfrac{y}{9}}^{2}}=1$

Thus, the equation of the locus of point P on the rod is ${{\dfrac{x}{81}}^{2}}+{{\dfrac{y}{9}}^{2}}=1$

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.

Ans: The given parabola is ${{x}^{2}}=12y$.

On comparing this equation with ${{x}^{2}}=4ay$, we’ll get $4a=12\Rightarrow a=3$

$\therefore$The coordinates of foci are S (0, a) = S (0, 3)

Let AB be the latus rectum of the given parabola.

The given parabola can be roughly drawn as

At $y=3,$ ${{x}^{2}}=12(3)\Rightarrow {{x}^{2}}=36\Rightarrow x=\pm 6$

$\therefore$The coordinates of A are (–6, 3), while the coordinates of B are (6, 3).

Therefore, the vertices of ΔOAB are O (0, 0), A (–6, 3), and B (6, 3).

Area of ΔOAB $=\dfrac{1}{2}\left| 0\left( 3-3 \right)+\left( -6 \right)\left( 3-0 \right)+6\left( 0-3 \right) \right|uni{{t}^{2}}$

$=\dfrac{1}{2}\left| \left( -6 \right)\left( 3 \right)+6\left( -3 \right) \right|uni{{t}^{2}}$

$=\dfrac{1}{2}\left| -18-18 \right|uni{{t}^{2}}$

$=\dfrac{1}{2}\left| -36 \right|uni{{t}^{2}}$

$=\dfrac{1}{2}\times 36uni{{t}^{2}}$

$=18$ $uni{{t}^{2}}$

Thus, the required area of the triangle is $18$ $uni{{t}^{2}}$.

7. A man running a racecourse notes that the sum of the distances from the two flag posts form 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.

Ans: Let A and B be the positions of the two flag posts and P(x, y) be the position of the man. Accordingly, PA + PB = 10.

We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is 10 m, while points A and B are the foci.

Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x-axis, the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form of $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1,$ where a is the semi-major axis.

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

Distance between the foci $(2c)=8\Rightarrow c=4$

On using the relation $c=\sqrt{{{a}^{2}}-{{b}^{2}}},$ we’ll get

$4=\sqrt{25-{{b}^{2}}}$

$\Rightarrow 16=25-{{b}^{2}}$

$\Rightarrow {{b}^{2}}=25-16$

$\Rightarrow {{b}^{2}}=9$

$\Rightarrow b=3$

Thus, the equation of the path traced by the man is $\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{9}=1$

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

Ans: Let OAB be the equilateral triangle inscribed in parabola ${{y}^{2}}=4ax$

Let AB intersect the x-axis at point C.

Let OC = k

From the equation of the given parabola, we have ${{y}^{2}}=4ak\Rightarrow y=\pm 2\sqrt{ak}$

$\therefore$The respective coordinates of points A and B are $\left( k,-2\sqrt{ak} \right)$ and $\left( k,-2\sqrt{ak} \right)$

AB = CA + CB = $2\sqrt{ak}+2\sqrt{ak}=4\sqrt{ak}$

Since OAB is an equilateral triangle, $O{{A}^{2}}=A{{B}^{2}}.$

$\therefore {{k}^{2}}+{{\left( 2\sqrt{ak} \right)}^{2}}={{\left( 4\sqrt{ak} \right)}^{2}}$

$\Rightarrow {{k}^{2}}+4ak=16ak$

$\Rightarrow {{k}^{2}}=12ak$

$\Rightarrow k=12a$

$\therefore AB=4\sqrt{ak}=4\sqrt{a\times 12a}$

$=4\sqrt{12{{a}^{2}}}$

$=8\sqrt{3}a$

Thus, the side of the equilateral triangle inscribed in parabola ${{y}^{2}}=4ax$ is $8\sqrt{3}a$

## Summary of Conic Sections

• A circle is the set of all points in a plane that are equidistant from a fixed point in the plane.

• The equation of a circle with centre $(h, k)$ and the radius $r$ is $(x-h)^2+(y-k)^2=r^2$

• A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point in the plane.

• The equation of the parabola with focus at $(a, 0) a>0$ and directrix $x=-a$ is $y^2=4 a x$.

• Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose endpoints lie on the parabola.

• Length of the latus rectum of the parabola $y^2=4 a x$ is $4 a$.

• An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant.

• The equation of an ellipse with foci on the $x$-axis is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

• Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose endpoints lie on the ellipse.

• Length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{2 b^2}{a}$.

• The eccentricity of an ellipse is the ratio between the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse.

• A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant.

• The equation of a hyperbola with foci on the $x$-axis is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

• Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola.

• Length of the latus rectum of the hyperbola: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{2 b^2}{a}$.

• The eccentricity of a hyperbola is the ratio of the distances from the centre of the hyperbola to one of the foci and to one of the vertices of the hyperbola.

## Overview of Deleted Syllabus for CBSE Class 11 Maths Chapter 10 Conic Sections

 Chapter Dropped Topics Conic Sections 11.5.2 - Special Cases of an Ellipse

## Class 11 Maths Chapter 10: Exercises Breakdown

 Exercise Number of Questions Exercise 10.1 15 Questions with Solutions Exercise 10.2 12 Questions with Solutions. Exercise 10.3 20 Questions with Solutions. Exercise 10.4 15 Questions with Solutions. Miscellaneous Exercise 8 Questions with Solutions.

## Conclusion

Chapter 10 - Conic Sections is crucial for mastering the fundamentals of ellipses, hyperbolas, and parabolas. It is important to focus on understanding their standard equations and properties. Practising the solved examples and exercises in the NCERT Solutions by Vedantu will help solidify these concepts. In the previous year's question paper, around 3 to 4 questions were asked from this chapter. Paying attention to these solutions will ensure you are well-prepared for your CBSE exams.

## Other Study Material for CBSE Class 11 Maths Chapter 10

 S.No Important Links for Chapter 10 Conic Sections 1 Class 11 Conic Sections Important Questions 2 Class 11 Conic Sections Revision Notes 3 Class 11 Conic Sections NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 11 Maths Chapter 10 Conic Sections

1. How many sections does Class 11 Maths Ch 10 cover?

Conic Section Class 11 Maths Ch 10 comprises some important sections that enhance your knowledge regarding topics like Hyperbola, Ellipse, Hyperbola etc. The chapter starts with an introductory part where you get an overall idea of what this chapter talks about.

The next few sections depict elements like degenerate conic sections, standard equations of Parabola, Ellipse, Latus Rectum, etc. You get the hold of these interesting topics which are used to solve tricky geometry questions. And in order to master these topics of Conic Sections Class 11, you can rely on NCERT solutions to help you with better methods.

2. How to get Full Marks in Conic Sections?

The Conic Sections chapter is an integral part of Class 11 Mathematics, and once you grasp the core areas of this chapter, it becomes easier to attain full marks. In order to understand the underlying areas, you should equally prioritize solving related problems to improve your skills.

Apart from that, most of the students struggle to find proper study materials that enhance their knowledge regarding a particular subject. The idea is to identify the weak areas and work on that by using proper resources and NCERT solutions perfectly assist with that.

3. Where can I get an Accurate Conic Section Class 11 Solution?

Accuracy is a vital factor when it comes to Mathematics; it requires lots of practice to achieve that. Since there are numerous options available on the internet, finding the right one becomes a little challenging.

However, you should opt for something that not only focuses on presenting advanced solutions but also an easy approach that is easy to comprehend.  NCERT solutions are solved by subject matter experts which makes it a reliable study material. Also, the simple formulas, detailed explanations and structured answers make it even better. The free PDF version can be easily downloaded from educational sites like Vedantu.

4. What are conic sections according to the Ch 10 Maths Class 11 Syllabus?

NCERT Class 11 Maths Chapter 10 comprises the topic Conic Sections, which explains further topics such as Ellipses, Circles, Parabolas and Hyperbolas. The chapter deals with different shapes, circular cones, semi-minor axis, the distance between focus and the central point of an ellipse. Conic Sections Class 11 solutions by Vedantu will help you develop a clear concept of this chapter as you go through it. These solutions contain the answers to all the questions from the NCERT textbook.

5. Which Chapter is Conic Sections in Class 11?

Ch 10 Maths Class 11 of the NCERT Maths textbook of Conic Sections. It mainly deals with various shapes, cones, and conic structures like parabolas, hyperbolas, circles, the axes, and the various sub topics related to these.

6. From where can I download the Conic Section Class 11 NCERT Solutions?

To download the Class 11 Chapter 10 NCERT Solutions, students can visit the page Conic Section Class 11 NCERT Solutions or visit Vedantu website. These solutions are designed by subject experts who have years of experience in this field. All the answers are given step-by-step, in simple language and are well-detailed, for the benefit of students. They can also refer to important questions, revision notes and other study materials available on Vedantu at free of cost, to help with their exam preparations. Students can also download the Vedantu app to access all these resources for free.

7. Do I need to practice all the questions provided in NCERT Solutions of Conic Section Class 11?

Students must practice all the questions provided in Class 11 Conic Section solution PDF, to score good marks in their exams. With this, they can understand all the concepts thoroughly and also increase their speed and accuracy of writing answers in the exam. These solutions can also come in handy when they get stuck while solving the questions from Chapter 10 of the Class 11 NCERT Maths textbook.

8. Is Conic Sections chapter important?

Conic Sections Class 11 Solutions is considered to be very important in Class 11, from the viewpoint of examinations. This chapter contains descriptions of certain mathematical figures like circles, hyperbolas, parabolas, and ellipses. These topics form a major part of the CBSE examination. This chapter is also important from the perspective of competitive exams like JEE Mains and JEE Advance. Students must be thorough with this chapter to ace both their school and competitive examinations.

9. What are the main topics covered in conic sections class 11 solutions?

Class 11 Conic Section solution pdf covers the fundamental concepts of conic sections, including the definitions and standard equations of ellipses, hyperbolas, and parabolas. It is important to understand how these shapes are derived and their unique properties. Focus on the derivations and applications as they form the basis of many exam questions.