NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4

Go back to  'Conic Sections'

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 \(\frac{{{y^2}}}{5} - \frac{{{x^2}}}{5} = 1\)

  
Download Cuemath NCERT App
Related Sections
Related Sections

Learn from the best math teachers and top your exams

Learn from the best

math teachers and top

your exams


Personalized Curriculum
Instant Doubts clarification
Cover latest CBSE Syllabus
Unlimited Mock & Practice tests
Covers CBSE, ICSE, IB curriculum

Instant doubt clearing with Cuemath Advanced Math Program
0