NCERT Solutions for Class 11 Maths Chapter 11: Conic Sections - Exercise 11.4

Exercise 11.4

Refer the page 23 to 39 for the exercise 11.4 in the PDF

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 obtain \[a = 4\] and \[b = 3\] , 

We know that 

\[{c^2} = {a^2} + {b^2}\]

Also, we know that \[a = 4\]  and \[b = 3\].

$\therefore {c^2} = {4^2} + {3^2}$

Put the value of $b$ and $a$ to determine the value of $c$.

${c^2} = 25$

$c = \sqrt {25} $

$c = 5$

Therefore,

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

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

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

As we know the value of $c$ and $a$. Thus,

$e = \dfrac{5}{4}$

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

So, $\dfrac{{2{b^2}}}{a} = \dfrac{9}{2}$

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 obtain \[a = 3\]and\[b = \sqrt {27} \]. 

We know that 

\[{c^2} = {b^2} + {a^2}\]

Put the value of $b$ and $a$ to determine the value of $c$.

$\therefore {c^2} = {3^2} + {\left( {\sqrt {27} } \right)^2}$

${c^2} = 9 + 27$

${c^2} = 36$

$c = \sqrt {36} $

$c = 6$

Therefore,

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

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

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

As we know the value of $c$ and $a$. Thus,

$e = \dfrac{6}{3}$

$e = 2$

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 

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

Dividing 36 in both side

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

$\dfrac{{{y^2}}}{{{2^2}}} - \dfrac{{{x^2}}}{{{3^2}}} = 1$

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

We know that,

\[{c^2} = {b^2} + {a^2}\]

$\therefore {c^2} = 4 + 9$

${c^2} = 13$

$c= \sqrt {13} $

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, $e = \dfrac{c}{a}$

As we know the value of $c$ and $a$. Thus,

$e = \dfrac{{\sqrt {13} }}{2}$

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 

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

Dividing by 576 in both side

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

$\dfrac{{{x^2}}}{{{6^2}}} - \dfrac{{{y^2}}}{{{8^2}}} = 1$

On comparing this equation with the standard equation of hyperbola i.e., $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$, we obtain \[a = 6\]and\[b = 8\] , 

We know that 

\[{c^2} = {b^2} + {a^2}\]

Put the value of $b$ and $a$ to determine the value of $c$.

$\therefore {c^2} = 36 + 64$

${c^2} = 100$

$c = 10$

Therefore,

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

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

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

As we know the value of $c$ and $a$. Thus,

$e = \dfrac{{10}}{6}$

$\dfrac{c}{a} = \dfrac{5}{3}$

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

So, $\dfrac{{2{b^2}}}{a} = \dfrac{{64}}{3}$

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 $5{y^2} - 9{x^2} = 36$. 

It can be written as 

$5{y^2} - 9{x^2} = 36$

Dividing 36 in both side and arrange the question similar with $\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$

$\dfrac{{{y^2}}}{{\left( {\dfrac{{36}}{5}} \right)}} - \dfrac{{{x^2}}}{4} = 1$

$\dfrac{{{y^2}}}{{{{\left( {\dfrac{6}{{\sqrt 5 }}} \right)}^2}}} - \dfrac{{{x^2}}}{{{2^2}}} = 1$………………………(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 obtain \[a = \dfrac{6}{{\sqrt 5 }}\] and\[b = 2\] ,

We know that 

\[{a^2} + {b^2} = {c^2}\]

Put the value of $b$ and $a$, also simplify the equation to determine the value of $c$.

$\therefore {c^2} = \dfrac{{36}}{5} + 4$

${c^2} = \dfrac{{36 + 20}}{5}$

${c^2} = \dfrac{{56}}{5}$

$c=\pm \sqrt{\frac{56}{5}}$

$c= 2\sqrt{\frac{14}{5}}$

Therefore,

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

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

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

As we know the value of $c$ and $a$. Thus,

$e = \dfrac{{\left( {\dfrac{{2\sqrt {14} }}{{\sqrt 5 }}} \right)}}{{\left( {\dfrac{6}{{\sqrt 5 }}} \right)}}$

$e = \dfrac{{2\sqrt {14} }}{6}$

$e = \dfrac{{\sqrt {14} }}{3}$

Length of latus rectum, $\dfrac{{2{b^2}}}{a} = \dfrac{{2 \times 4}}{{\dfrac{6}{{\sqrt 5 }}}} = \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 

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

By dividing 784 in both sides, we get,

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

$\dfrac{{{y^2}}}{{{4^2}}} - \dfrac{{{x^2}}}{{{7^2}}} = 1$

On comparing this equation with the standard equation of hyperbola i.e., $\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$, we obtain \[a = 4\] and \[b = 7\]. 

We know that 

\[{c^2} = {b^2} + {a^2}\]

Put the value of $b$ and $a$ and simplify the equation to determine the value of $c$.

$\therefore {c^2} = 16 + 49$

${c^2} = 65$

$c= \sqrt{65}$

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, $e = \dfrac{c}{a}$

As we know the value of $c$ and $a$. Thus,

$e = \dfrac{{\sqrt {65} }}{4}$

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

So, $\dfrac{{2{b^2}}}{a} = \dfrac{{49}}{2}$.

7. Find the equation of the hyperbola satisfying the given conditions: Vertices \[\left( { \pm {\mathbf{2}},{\mathbf{0}}} \right),\]foci \[\left( { \pm {\mathbf{3}},{\mathbf{0}}} \right)\].

Ans: Vertices \[\left( { \pm {\mathbf{2}},{\mathbf{0}}} \right)\] , foci \[\left( { \pm {\mathbf{3}},{\mathbf{0}}} \right)\]

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 \[\left( { \pm 2,0} \right)\], so $a = 2$. 

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

We know that

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

Put the value of $a$ and $c$ and simplify the equation to determine the value of ${b^2}$.

$\therefore {2^2} + {b^2} = {3^2}$

\[{b^2} = 9 - 4\]

${b^2} = 5$

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 given conditions: Vertices \[\left( {{\mathbf{0}}, \pm {\mathbf{5}}} \right)\], foci \[\left( {{\mathbf{0}}, \pm {\mathbf{8}}} \right)\]

Ans:  Vertices \[\left( {{\mathbf{0}}, \pm {\mathbf{5}}} \right)\] , foci \[\left( {{\mathbf{0}}, \pm 8} \right)\]

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 \[\left( {{\mathbf{0}}, \pm {\mathbf{5}}} \right)\] , so $a = 5$. 

Since the foci are \[\left( {{\mathbf{0}}, \pm 8} \right)\] , so $c = 8$. 

We know that 

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

Put the value of $a$ and $c$ to get the value of $b$,

$\therefore {5^2} + {b^2} = {8^2}$

${b^2} = {8^2} - {5^2}$

\[{b^2} = 64 - 25\]

${b^2} = 39$

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 given conditions: Vertices \[\left( {{\mathbf{0}}, \pm 3} \right)\] , foci \[\left( {{\mathbf{0}}, \pm 5} \right)\]

Ans:  Vertices \[\left( {{\mathbf{0}}, \pm 3} \right)\] , foci \[\left( {{\mathbf{0}}, \pm 5} \right)\]

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 \[\left( {{\mathbf{0}}, \pm 3} \right)\] , so $a = 3$. 

Since the foci are \[\left( {{\mathbf{0}}, \pm 5} \right)\] , so $c = 5$. 

We know that 

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

Put the value of $c$ and $a$ and simplify the equation to determine the value of ${b^2}$.

$ \therefore {3^2} + {b^2} $ = 25

\[{b^2} = 25 - 9\]

${b^2} = 16$

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 given conditions: Foci \[\left( { \pm 5,0} \right)\] , the transverse axis is of length 8. 

Ans: 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 $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ . 

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

Since the length of the transverse axis is 8,

\[2a = 8\]

Dividing both side with 2

\[a = 4\]

We know that 

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

Put the value of $c$ and $a$ and simplify the equation to determine the value of $b$.

$ \therefore {4^2} + {b^2} $ = 25

\[{b^2} = 25 - 16\]

${b^2} = 9$

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 given conditions: Foci \[\left( {{\mathbf{0}}, \pm 13} \right)\] , the conjugate axis is of length 24. 

Ans: Foci \[\left( {{\mathbf{0}}, \pm 13} \right)\] , & 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 $\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$ . 

Since the foci are \[\left( {{\mathbf{0}}, \pm 13} \right)\] , so  $c = 13$. 

Since the length of the  conjugate axis  is 24,

\[2b = 24\]

Dividing 2 on both sides,

\[b = 12\]

We know that 

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

Put the value of $b$ and $c$, also simplify the equation to determine the value of $a$.

\[\therefore {a^2} + {12^2} = {13^2}\]

\[{a^2} = 169 - 144\]

${a^2} = 25$

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 given 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 \[\left( { \pm 3\sqrt 5 ,0} \right)\] , so  $c =  \pm 3\sqrt 5 $. 

Since the length of the latus rectum is 8,

\[\dfrac{{2{b^2}}}{a} = 8\]

Multiplying both sides by $\dfrac{a}{2}$,

\[{b^2} = \dfrac{{8a}}{2}\]

\[{b^2} = 4a\]

We know that 

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

Put the value of $b$ and $c$,  simplify the equation to determine the value of $a$.

\[\therefore {a^2} + 4a = 45\]

\[{a^2} + 4a - 45 = 0\]

\[{a^2} + 9a - 5a - 45 = 0\]

$\left( {a + 9} \right)\left( {a - 5} \right) = 0$

$a =  - 9, 5$

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 given 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 \[\left( { \pm 4,0} \right)\] , so  $c = 4$. 

Since the latus rectum is of length 12. 

\[\dfrac{{2{b^2}}}{a} = 12\]

Multiply both sides with $\dfrac{a}{2}$,

\[{b^2} = 6a\]…………….(1)

We know that 

${a^2} + {b^2} = {c^2}$…………..(2)

Merge the equation (1) and (2) and simplify the equation to get the value of $a$,

\[\therefore {a^2} + 6a = 16\]

\[{a^2} + 6a - 16 = 0\]

\[{a^2} + 8a - 2a - 16 = 0\]

\[\left( {a + 8} \right)\left( {a - 2} \right) = 0\]

\[a =  - 8, 2\]

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 given 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 \[\left( { \pm 7,0} \right)\] , so $a = 7$. 

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

$\therefore \dfrac{c}{a} = \dfrac{4}{3} $

$\dfrac{c}{7} = \dfrac{4}{3}$

Multiply both sides with 7

$c = \dfrac{{28}}{3}$

We know that 

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

Put the value of $c$ and $a$ ,  also simplify the equation to determine the value of $b$.

\[\therefore {7^2} + {b^2} = {\left( {\dfrac{{28}}{3}} \right)^2}\]

${b^2} = \dfrac{{784}}{9} - 49$

${b^2} = \dfrac{{784 - 441}}{9}$

${b^2} = \dfrac{{343}}{9}$

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 \[\left( {{\mathbf{0}}, \pm \sqrt {10} } \right)\] , passing through \[\left( {{\mathbf{2}},{\mathbf{3}}} \right)\]

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

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 \[\left( {{\mathbf{0}}, \pm \sqrt {10} } \right)\] , so $c = \sqrt {10} $. 

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

Put the value of $c$,

\[\therefore {a^2} + {b^2} = 10\]

\[{b^2} = 10 - {a^2}\]………..… (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 obtain the value of ${a^2}$

$\dfrac{9}{{{a^2}}} - \dfrac{4}{{\left( {10 - {a^2}} \right)}} = 1$

$9(10 - {a^2}) - 4{a^2} = {a^2}\left( {10 - {a^2}} \right)$

$90 - 9{a^2} - 4{a^2} = 10{a^2} - {a^4}$

${a^4} - 23{a^2} + 90 = 0$

${a^4} - 18{a^2} - 5{a^2} + 90 = 0$

${a^2}\left( {{a^2} - 18} \right) - 5\left( {{a^2} - 18} \right) = 0$

\[{a^2} = 18,5\]

In hyperbola,

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

$\therefore {a^2} = 5$

Since the value of $a$ is positive as $e>1$ so ${b^2} = 10 - {a^2}$

As we know the value of $a$,

${b^2} = 10 - 5 = 5$

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

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.4

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