NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.3


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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