# NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.3

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## Chapter 11 Ex.11.3 Question 1

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $$\frac{{x^2}}{{36}} + \frac{{y^2}}{16} = 1$$

### Solution

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

Here, the denominator of $$\frac{{x^2}}{36}$$  is greaterthan the denominator of $$\frac{{y^2}}{16}$$.

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

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

Hence,

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

Therefore,

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

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

Length of major axis $$= 2a = 12$$

Length of minor axis $$= 2b = 8$$

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

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

## Chapter 11 Ex.11.3 Question 2

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{25}} = 1$$

### Solution

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

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

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

On comparing the given equation with $$\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1$$, we obtain $$b = 2$$ and $$a = 5$$

Hence,

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

Therefore,

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

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

Length of major axis $$= 2a = 10$$

Length of minor axis $$= 2b = 4$$

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

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

## Chapter 11 Ex.11.3 Question 3

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $$\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$$

### Solution

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

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

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

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

Hence,

\begin{align}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 $$\left( { \pm 4,0} \right)$$

Length of major axis $$= 2a = 8$$

Length of minor axis $$= 2b = 6$$

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

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

## Chapter 11 Ex.11.3 Question 4

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{100}} = 1$$

### Solution

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

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

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

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

Hence,

\begin{align}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 $$\left( {0, \pm 10} \right)$$

Length of major axis $$= 2a = 20$$

Length of minor axis $$= 2b = 10$$

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

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

## Chapter 11 Ex.11.3 Question 5

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $$\frac{{{x^2}}}{{49}} + \frac{{{y^2}}}{{36}} = 1$$

### Solution

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

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

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

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

Hence,

\begin{align}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 $$\left( { \pm 7,\;0} \right)$$

Length of major axis $$= 2a = 14$$

Length of minor axis $$= 2b = 12$$

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

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

## Chapter 11 Ex.11.3 Question 6

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $$\frac{{x^2}}{100} + \frac{{y^2}}{400} = 1$$

### Solution

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

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

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

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

Hence,

\begin{align}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 $$\left( {0, \pm 20} \right)$$

Length of major axis $$= 2a = 40$$

Length of minor axis $$= 2b = 20$$

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

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

## Chapter 11 Ex.11.3 Question 7

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $$36{x^2} + 4{y^2} = 144$$

### Solution

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

It can be written as,

\begin{align}&36{x^2} + 4{y^2} = 144\\&\Rightarrow\; \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{36}} = 1\\&\Rightarrow \;\frac{{{x^2}}}{{{2^2}}} + \frac{{{y^2}}}{{{6^2}}} = 1 \qquad \ldots \left( 1 \right)\end{align}

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

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

On comparing equation$$\left( 1 \right)$$with $$\frac{{x^2}}{{b^2}} + \frac{{y^2}}{{a^2}} = 1$$, we obtain $$a = 6$$ and $$b = 2$$

Hence,

\begin{align}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 $$\left( {0, \pm 6} \right)$$

Length of major axis $$= 2a = 12$$

Length of minor axis $$= 2b = 4$$

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

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

## Chapter 11 Ex.11.3 Question 8

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $$16{x^2} + {y^2} = 16$$

### Solution

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

It can be written as,

\begin{align}&16{x^2} + {y^2} = 16\\ &\Rightarrow \; \;\frac{{{x^2}}}{1} + \frac{{{y^2}}}{{16}} = 1\\ & \Rightarrow \; \;\frac{{{x^2}}}{{{1^2}}} + \frac{{{y^2}}}{{{4^2}}} = 1 \qquad \quad \ldots \left( 1 \right) \end{align}

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

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

On comparing equation $$\left( 1 \right)$$ with $$\frac{{x^2}}{{b^2}} + \frac{{y^2}}{{a^2}} = 1$$ , we obtain $$a = 4$$ and $$b = 1$$

Hence,

\begin{align}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 $$\left( {0, \pm 4} \right)$$

Length of major axis $$= 2a = 8$$

Length of minor axis $$= 2b = 2$$

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

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

## Chapter 11 Ex.11.3 Question 9

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $$4{x^2} + 9{y^2} = 36$$

### Solution

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

It can be written as,

\begin{align}&4{x^2} + 9{y^2} = 36\\&\Rightarrow\; \frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1\\&\Rightarrow \;\frac{{{x^2}}}{{{3^2}}} + \frac{{{y^2}}}{{{2^2}}} = 1 \qquad \quad \ldots \left( 1 \right)\end{align}

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

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

On comparing equation $$\left( 1 \right)$$ with $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$, we obtain $$a = 3$$ and $$b = 2$$

Hence,

\begin{align}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 $$\left( { \pm 3,0} \right)$$

Length of major axis $$= 2a = 6$$

Length of minor axis $$= 2b = 4$$

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

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

## Chapter 11 Ex.11.3 Question 10

Find the equation for the ellipse that satisfies the given conditions: Vertices $$\left( { \pm 5,0} \right)$$, Foci$$\left( { \pm 4,0} \right)$$.

### Solution

Vertices $$\left( { \pm 5,0} \right)$$, Foci $$\left( { \pm 4,0} \right)$$

Here, the vertices are on the $$x$$-axis.

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

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

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

Hence,

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

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

## Chapter 11 Ex.11.3 Question 11

Find the equation for the ellipse that satisfies the given conditions: Vertices $$\left( {0, \pm 13} \right)$$, Foci$$\left( {0, \pm 5} \right)$$

### Solution

Vertices $$\left( {0, \pm 13} \right)$$, Foci $$\left( {0, \pm 5} \right)$$

Here, the vertices are on the $$y$$-axis.

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

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

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

Hence,

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

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

## Chapter 11 Ex.11.3 Question 12

Find the equation for the ellipse that satisfies the given conditions: Vertices $$\left( { \pm 6,0} \right)$$, Foci$$\left( { \pm 4,0} \right)$$.

### Solution

Vertices $$\left( { \pm 6,0} \right)$$, Foci $$\left( { \pm 4,0} \right)$$

Here, the vertices are on the $$x -$$axis.

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

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

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

Hence,

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

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

## Chapter 11 Ex.11.3 Question 13

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis$$\left( { \pm 3,0} \right)$$, Ends of minor axis $$\left( {0, \pm 2} \right)$$

### Solution

Ends of major axis $$\left( { \pm 3,0} \right)$$, Ends of minor axis $$\left( 0,\pm 2 \right)$$

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

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

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

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

## Chapter 11 Ex.11.3 Question 14

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

### Solution

Ends of major axis $$\left( {0, \pm \sqrt 5 } \right)$$, Ends of minor axis $$\left( { \pm 1,0} \right)$$

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

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

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

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

## Chapter 11 Ex.11.3 Question 15

Find the equation for the ellipse that satisfies the given conditions: Length of major axis $$26$$, Foci $$\left( { \pm 5,0} \right)$$

### Solution

Length of major axis $$= 26$$, Foci $$= \left( { \pm 5,0} \right)$$

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 $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$, where $$a$$ is the semi major axis.

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

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

Hence,

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

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

## Chapter 11 Ex.11.3 Question 16

Find the equation for the ellipse that satisfies the given conditions: Length of minor axis $$16$$, Foci $$\left( {0, \pm 6} \right)$$

### Solution

Length of minor axis $$= 16$$, Foci $$= \left( {0, \pm 6} \right)$$

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 $$\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1$$, where $$a$$ is the semi major axis.

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

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

Hence,

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

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

## Chapter 11 Ex.11.3 Question 17

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

### Solution

Foci $$\left( { \pm 3,0} \right)$$, $$a = 4$$

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 $$\frac{{x^2}}{{a^2}} + \frac{{y^2}}{{b^2}} = 1$$, where $$a$$ is the semi major axis.

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

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

Hence,

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

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

## Chapter 11 Ex.11.3 Question 18

Find the equation for the ellipse that satisfies the given conditions: $$b = 3,\;c = 4$$, centre at the origin; foci on the ­$$x$$-axis.

### Solution

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 $$\frac{{x^2}}{{a^2}} + \frac{{y^2}}{{b^2}} = 1$$, where $$a$$ is the semi major axis.

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

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

Hence,

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

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

## Chapter 11 Ex.11.3 Question 19

Find the equation for the ellipse that satisfies the given conditions: Centre at $$\left( {0,0} \right)$$, major axis on the $$y$$-axis and passes through the points $$\left( {3,2} \right)$$ and $$\left( {1,6} \right)$$.

### Solution

Since the centre is at $$\left( {0,0} \right)$$ and the major axis is on the $$y$$-axis, the equation of the ellipse will be of the form $$\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1$$, where $$a$$ is the semi major axis.

The ellipse passes through points $$\left( {3,2} \right)$$ and $$\left( {1,6} \right)$$

Hence,

\begin{align}\frac{9}{{{b^2}}} + \frac{4}{{{a^2}}} &= 1 \qquad \quad \ldots \left( 1 \right)\\\frac{1}{{{b^2}}} + \frac{{36}}{{{a^2}}} &= 1\qquad \quad \ldots \left( 2 \right)\end{align}

On solving equations $$\left( 1 \right)$$ and $$\left( 2 \right)$$, we obtain $${a^2} = 40$$ and $${b^2} = 10$$.

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

## Chapter 11 Ex.11.3 Question 20

Find the equation for the ellipse that satisfies the given conditions: Major axis on the $$x$$-axis and passes through the points $$\left( {4,3} \right)$$ and $$\left( {6,2} \right)$$.

### Solution

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

The ellipse passes through points $$\left( 4,3 \right)$$ and $$\left( 6,2 \right)$$.

Hence,

\begin{align}\frac{{16}}{{{b^2}}} + \frac{9}{{{a^2}}} = 1 \qquad \ldots \left( 1 \right)\\\frac{{36}}{{{b^2}}} + \frac{4}{{{a^2}}} = 1 \qquad \ldots \left( 2 \right)\end{align}

On solving equations $$\left( 1 \right)$$ and $$\left( 2 \right)$$, we obtain $${a^2} = 52$$ and $${b^2} = 13$$.

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

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