NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4

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Chapter 11 Ex.11.4 Question 1

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

Solution

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

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

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

Hence,

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

Therefore,

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

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

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

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

Chapter 11 Ex.11.4 Question 2

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

Solution

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

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

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

Hence,

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

Therefore,

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

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

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

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

Chapter 11 Ex.11.4 Question 3

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

Solution

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

It can be written as

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

On comparing equation $$\left( 1 \right)$$ with the standard equation of hyperbola, i.e., $$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$, we obtain $$a = 2$$ and $$b = 3$$

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

Hence,

\begin{align}&\Rightarrow\; {c^2} = {2^2} + {3^2}\\&\Rightarrow \;{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 $$\left( {0, \pm 2} \right)$$

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

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

Chapter 11 Ex.11.4 Question 4

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

Solution

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

It can be written as

\begin{align}&16{x^2} - 9{y^2} = 576\\&\Rightarrow \;\frac{x^2}{36} - \frac{y^2}{64} = 1\\&\Rightarrow \;\frac{x^2}{6^2} - \frac{y^2}{8^2} = 1 \qquad \ldots \left( 1 \right)\end{align}

On comparing equation $$\left( 1 \right)$$ with the standard equation of hyperbola, i.e., $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, we obtain $$a = 6$$ and $$b = 8$$

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

Hence,

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

Therefore,

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

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

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

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

Chapter 11 Ex.11.4 Question 5

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

Solution

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

It can be written as

\begin{align}&5{y^2} - 9{x^2} = 36\\&\Rightarrow \;\frac{y^2}{{\left( {\frac{36}{5}} \right)}} - \frac{{x^2}}{4} = 1\\&\Rightarrow \;\frac{{y^2}}{{\left( {\frac{6}{\sqrt 5 }} \right)}} - \frac{{x^2}}{{2^2}} = 1 \qquad \ldots \left( 1 \right)\end{align}

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

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

Hence,

\begin{align}&\Rightarrow \;{c^2} = {\left( {\frac{6}{{\sqrt 5 }}} \right)^2} + {\left( 2 \right)^2}\\&\Rightarrow \;{c^2} = \frac{{36}}{5} + 4\\&\Rightarrow \;{c^2} = \frac{{56}}{5}\\&\Rightarrow \;c = \sqrt {\frac{{56}}{5}} \\&\Rightarrow\; c = \frac{{2\sqrt {14} }}{{\sqrt 5 }}\end{align}

Therefore,

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

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

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

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

Chapter 11 Ex.11.4 Question 6

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

Solution

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

It can be written as $$49{y^2} - 16{x^2} = 784$$

\begin{align}&49{y^2} - 16{x^2} = 784\\&\Rightarrow \;\frac{{y^2}}{{16}} - \frac{{x^2}}{{49}} = 1\\&\Rightarrow \;\frac{{y^2}}{{4^2}} - \frac{{x^2}}{{7^2}} = 1 \qquad \ldots \left( 1 \right)\end{align}

On comparing equation $$\left( 1 \right)$$ with the standard equation of hyperbola, i.e., $$\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1$$, we obtain $$a = 4$$ and $$b = 7$$

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

Hence,

\begin{align}&\Rightarrow \;{c^2} = {\left( 4 \right)^2} + {\left( 7 \right)^2}\\&\Rightarrow\; {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 $$\left( {0, \pm 4} \right)$$

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

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

Chapter 11 Ex.11.4 Question 7

Find the equation of the hyperbola satisfying the given conditions: Vertices $$\left( { \pm 2,0} \right)$$, Foci$$\left( { \pm 3,0} \right)$$

Solution

Vertices $$\left( { \pm 2,0} \right)$$, Foci $$\left( { \pm 3,0} \right)$$

Here, the vertices are on the ix-axis.

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

Since the vertices are $$\left( { \pm 2,0} \right)$$, $$a = 2$$

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

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

Hence,

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

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

Chapter 11 Ex.11.4 Question 8

Find the equation of the hyperbola satisfying the given conditions: Vertices $$\left( {0, \pm 5} \right)$$, Foci $$\left( {0, \pm 8} \right)$$

Solution

Vertices $$\left( {0, \pm 5} \right)$$, Foci $$\left( {0, \pm 8} \right)$$

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

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

Since the vertices are $$\left( {0, \pm 5} \right)$$, $$a = 5$$

Since the foci are $$\left( {0, \pm 8} \right)$$, $$c = 8$$

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

Hence,

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

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

Chapter 11 Ex.11.4 Question 9

Find the equation of the hyperbola satisfying the given conditions: Vertices $$\left( {0, \pm 3} \right)$$, Foci $$\left( {0, \pm 5} \right)$$

Solution

Vertices $$\left( {0, \pm 3} \right)$$, Foci $$\left( {0, \pm 5} \right)$$

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

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

Since the vertices are $$\left( {0, \pm 3} \right)$$, $$a = 3$$

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

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

Hence,

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

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

Chapter 11 Ex.11.4 Question 10

Find the equation of the hyperbola satisfying the given conditions: Foci $$\left( { \pm 5,0} \right)$$, the transverse axis is of length $$8.$$

Solution

Foci $$\left( { \pm 5,0} \right)$$, 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 $$\frac{{x^2}}{{a^2}} - \frac{{y^2}}{{b^2}} = 1$$

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

Since the length of the transverse axis is $$8,$$

Then,

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

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

Hence,

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

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

Chapter 11 Ex.11.4 Question 11

Find the equation of the hyperbola satisfying the given conditions: Foci $$\left( {0, \pm 13} \right)$$, the conjugate axis is of length $$24.$$

Solution

Foci $$\left( {0, \pm 13} \right)$$, the conjugate axis is of length $$24$$

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

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

Since the foci are $$\left( {0, \pm 13} \right)$$, $$c = 13$$

Since the length of the transverse axis is $$24$$,

Then,

\begin{align}&\Rightarrow\; 2b = 24\\&\Rightarrow\; b = 12\end{align}

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

Hence,

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

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

Chapter 11 Ex.11.4 Question 12

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

Solution

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

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

Since the length of latus rectum is $$8$$

Then,

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

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

Hence,

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

\begin{align}&\Rightarrow\; {b^2} = 4a\\&\Rightarrow\; {b^2} = 4 \times 5\\&\Rightarrow \;{b^2} = 20\end{align}

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

Chapter 11 Ex.11.4 Question 13

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

Solution

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

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

Since the length of latus rectum is $$12$$

Then,

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

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

Hence,

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

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

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

Chapter 11 Ex.11.4 Question 14

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

Solution

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

Here, the foci are on the x-axis.

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

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

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

Hence,

\begin{align}&\Rightarrow e = \frac{c}{a} = \frac{4}{3}\\&\Rightarrow \frac{c}{7} = \frac{4}{3}\\&\Rightarrow c = \frac{{28}}{3}\end{align}

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

Therefore,

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

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

Chapter 11 Ex.11.4 Question 15

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

Solution

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

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

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

Hence,

\begin{align}&\Rightarrow {a^2} + {b^2} = 10\\&\Rightarrow {b^2} = 10 - {a^2}\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

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

$\frac{9}{{{a^2}}} - \frac{4}{{{b^2}}} = 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)$

From equations $$\left( 1 \right)$$ and $$\left( 2 \right)$$, we obtain

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

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

Therefore,

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

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

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