NCERT Solutions for Exercise 11.4 Class 11 Maths Chapter 11 - Conic Section

Conic Section Class 11 Chapter 11 Exercise 11.4

Question:1 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

$\frac{x^2}{16} - \frac{y^2}{9} = 1$

Answer:

Given a Hyperbola equation,

$\frac{x^2}{16} - \frac{y^2}{9} = 1$

Can also be written as

$\frac{x^2}{4^2} - \frac{y^2}{3^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

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

We get,

$a=4$ and $b=3$

Now, As we know the relation in a hyperbola,

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

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

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

$c=5$

Here as we can see from the equation that the axis of the hyperbola is X -axis. So,

Coordinates of the foci:

$(c,0) \:and\:(-c,0)=(5,0)\:and\:(-5,0)$

The Coordinates of vertices:

$(a,0) \:and\:(-a,0)=(4,0)\:and\:(-4,0)$

The Eccentricity:

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

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(3)^2}{4}=\frac{18}{4}=\frac{9}{2}$

Question:2Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

$\frac{y^2}{9} - \frac{x^2}{27} = 1$

Answer:

Given a Hyperbola equation,

$\frac{y^2}{9} - \frac{x^2}{27} = 1$

Can also be written as

$\frac{y^2}{3^2} - \frac{x^2}{(\sqrt{27})^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

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

We get,

$a=3$ and $b=\sqrt{27}$

Now, As we know the relation in a hyperbola,

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

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

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

$c=\sqrt{36}$

$c=6$

Here as we can see from the equation that the axis of the hyperbola is Y-axis. So,

Coordinates of the foci:

$(0,c) \:and\:(0,-c)=(0,6)\:and\:(0,-6)$

The Coordinates of vertices:

$(0,a) \:and\:(0,-a)=(0,3)\:and\:(0,-3)$

The Eccentricity:

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

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(27)}{3}=\frac{54}{3}=18$

Question:3 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

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

Answer:

Given a Hyperbola equation,

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

Can also be written as

$\frac{9y^2}{36} - \frac{4x^2}{36} = 1$

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

Comparing this equation with the standard equation of the hyperbola:

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

We get,

$a=2$ and $b=3$

Now, As we know the relation in a hyperbola,

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

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

$c=\sqrt{2^2+3^2}$

$c=\sqrt{13}$

Hence,

Coordinates of the foci:

$(0,c) \:and\:(0,-c)=(0,\sqrt{13})\:and\:(0,-\sqrt{13})$

The Coordinates of vertices:

$(0,a) \:and\:(0,-a)=(0,2)\:and\:(0,-2)$

The Eccentricity:

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

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(9)}{2}=\frac{18}{2}=9$

Question:4 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

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

Answer:

Given a Hyperbola equation,

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

Can also be written as

$\frac{16x^2}{576} - \frac{9y^2}{576} = 1$

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

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

Comparing this equation with the standard equation of the hyperbola:

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

We get,

$a=6$ and $b=8$

Now, As we know the relation in a hyperbola,

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

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

$c=\sqrt{6^2+8^2}$

$c=10$

Therefore,

Coordinates of the foci:

$(c,0) \:and\:(-c,0)=(10,0)\:and\:(-10,0)$

The Coordinates of vertices:

$(a,0) \:and\:(-a,0)=(6,0)\:and\:(-6,0)$

The Eccentricity:

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

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(8)^2}{6}=\frac{128}{6}=\frac{64}{3}$

Question:5 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

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

Answer:

Given a Hyperbola equation,

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

Can also be written as

$\frac{5y^2}{36} - \frac{9x^2}{36} = 1$

$\frac{y^2}{\frac{36}{5}} - \frac{x^2}{4} = 1$

$\frac{y^2}{(\frac{6}{\sqrt{5}})^2} - \frac{x^2}{2^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

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

We get,

$a=\frac{6}{\sqrt{5}}$

and $b=2$

Now, As we know the relation in a hyperbola,

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

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

$c=\sqrt{(\frac{6}{\sqrt{5}})^2+2^2}$

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

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

Here as we can see from the equation that the axis of the hyperbola is Y-axis. So,

Coordinates of the foci:

$(0,c) \:and\:(0,-c)=\left(0,2\sqrt{\frac{14}{5}}\right)\:and\:\left(0,-2\sqrt{\frac{14}{5}}\right)$

The Coordinates of vertices:

$(0,a) \:and\:(0,-a)=\left(0,\frac{6}{\sqrt{5}}\right)\:and\:\left(0,-\frac{6}{\sqrt{5}}\right)$

The Eccentricity:

$e=\frac{c}{a}=\frac{2\sqrt{\frac{14}{5}}}{\frac{6}{\sqrt{5}}}=\frac{\sqrt{14}}{3}$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(4)}{\frac{6}{\sqrt{5}}}=\frac{4\sqrt{5}}{3}$

Question:6 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

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

Answer:

Given a Hyperbola equation,

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

Can also be written as

$\frac{49y^2}{784} - \frac{16x^2}{784} = 1$

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

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

Comparing this equation with the standard equation of the hyperbola:

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

We get,

$a=4$ and $b=7$

Now, As we know the relation in a hyperbola,

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

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

$c=\sqrt{4^2+7^2}$

$c=\sqrt{65}$

Therefore,

Coordinates of the foci:

$(0,c) \:and\:(0,-c)=(0,\sqrt{65})\:and\:(0,-\sqrt{65})$

The Coordinates of vertices:

$(0,a) \:and\:(0,-a)=(0,4)\:and\:(0,-4)$

The Eccentricity:

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

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(49)}{4}=\frac{98}{4}=\frac{49}{2}$

Question:7 Find the equations of the hyperbola satisfying the given conditions.

Vertices (± 2, 0), foci (± 3, 0)

Answer:

Given, in a hyperbola

Vertices (± 2, 0), foci (± 3, 0)

Here, Vertices and focii are on the X-axis so, the standard equation of the Hyperbola will be ;

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

By comparing the standard parameter (Vertices and foci) with the given one, we get

$a=2$ and $c=3$

Now, As we know the relation in a hyperbola

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

$b^2=c^2-a^2$

$b^2=3^2-2^2$

$b^2=9-4=5$

Hence,The Equation of the hyperbola is ;

$\frac{x^2}{4}-\frac{y^2}{5}=1$

Question:8 Find the equations of the hyperbola satisfying the given conditions.

Vertices (0, ± 5), foci (0, ± 8)

Answer:

Given, in a hyperbola

Vertices (0, ± 5), foci (0, ± 8)

Here, Vertices and focii are on the Y-axis so, the standard equation of the Hyperbola will be ;

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

By comparing the standard parameter (Vertices and foci) with the given one, we get

$a=5$ and $c=8$

Now, As we know the relation in a hyperbola

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

$b^2=c^2-a^2$

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

$b^2=64-25=39$

Hence, The Equation of the hyperbola is ;

$\frac{y^2}{25}-\frac{x^2}{39}=1$.

Question:9 Find the equations of the hyperbola satisfying the given conditions.

Vertices (0, ± 3), foci (0, ± 5)

Answer:

Given, in a hyperbola

Vertices (0, ± 3), foci (0, ± 5)

Here, Vertices and focii are on the Y-axis so, the standard equation of the Hyperbola will be ;

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

By comparing the standard parameter (Vertices and foci) with the given one, we get

$a=3$ and $c=5$

Now, As we know the relation in a hyperbola

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

$b^2=c^2-a^2$

$b^2=5^2-3^2$

$b^2=25-9=16$

Hence, The Equation of the hyperbola is ;

$\frac{y^2}{9}-\frac{x^2}{16}=1$.

Question:10 Find the equations of the hyperbola satisfying the given conditions.

Foci (± 5, 0), the transverse axis is of length 8.

Answer:

Given, in a hyperbola

Foci (± 5, 0), the transverse axis is of length 8.

Here, focii are on the X-axis so, the standard equation of the Hyperbola will be ;

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

By comparing the standard parameter (transverse axis length and foci) with the given one, we get

$2a=8\Rightarrow a=4$ and $c=5$

Now, As we know the relation in a hyperbola

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

$b^2=c^2-a^2$

$b^2=5^2-4^2$

$b^2=25-16=9$

Hence, The Equation of the hyperbola is ;

$\frac{x^2}{16}-\frac{y^2}{9}=1$

Question:11 Find the equations of the hyperbola satisfying the given conditions.

Foci (0, ±13), the conjugate axis is of length 24.

Answer:

Given, in a hyperbola

Foci (0, ±13), the conjugate axis is of length 24.

Here, focii are on the Y-axis so, the standard equation of the Hyperbola will be ;

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

By comparing the standard parameter (length of conjugate axis and foci) with the given one, we get

$2b=24\Rightarrow b=12$ and $c=13$

Now, As we know the relation in a hyperbola

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

$a^2=c^2-b^2$

$a^2=13^2-12^2$

$a^2=169-144=25$

Hence, The Equation of the hyperbola is ;

$\frac{y^2}{25}-\frac{x^2}{144}=1$.

Question:12 Find the equations of the hyperbola satisfying the given conditions.

Foci $(\pm 3\sqrt5, 0)$, the latus rectum is of length 8.

Answer:

Given, in a hyperbola

Foci $(\pm 3\sqrt5, 0)$, the latus rectum is of length 8.

Here, focii are on the X-axis so, the standard equation of the Hyperbola will be ;

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

By comparing standard parameter (length of latus rectum and foci) with the given one, we get

$c=3\sqrt{5}$ and

$\frac{2b^2}{a}=8\Rightarrow 2b^2=8a\Rightarrow b^2=4a$

Now, As we know the relation in a hyperbola

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

$c^2=a^2+4a$

$a^2+4a=(3\sqrt{5})^2$

$a^2+4a=45$

$a^2+9a-5a-45=0$

$(a+9)(a-5)=0$

$a=-9\:or\:5$

Since $a$ can never be negative,

$a=5$

$a^2=25$

$b^2=4a=4(5)=20$

Hence, The Equation of the hyperbola is ;

$\frac{x^2}{25}-\frac{y^2}{20}=1$

Question:13 Find the equations of the hyperbola satisfying the given conditions.

Foci (± 4, 0), the latus rectum is of length 12

Answer:

Given, in a hyperbola

Foci (± 4, 0), the latus rectum is of length 12

Here, focii are on the X-axis so, the standard equation of the Hyperbola will be ;

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

By comparing standard parameter (length of latus rectum and foci) with the given one, we get

$c=4$ and

$\frac{2b^2}{a}=12\Rightarrow 2b^2=12a\Rightarrow b^2=6a$

Now, As we know the relation in a hyperbola

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

$c^2=a^2+6a$

$a^2+6a=4^2$

$a^2+6a=16$

$a^2+8a-2a-16=0$

$(a+8)(a-2)=0$

$a=-8\:or\:2$

Since $a$ can never be negative,

$a=2$

$a^2=4$

$b^2=6a=6(2)=12$

Hence, The Equation of the hyperbola is ;

$\frac{x^2}{4}-\frac{y^2}{12}=1$

Question:14 Find the equations of the hyperbola satisfying the given conditions.

vertices (± 7,0), $e = \frac{4}{3}$

Answer:

Given, in a hyperbola

vertices (± 7,0), And

$e = \frac{4}{3}$

Here, Vertices is on the X-axis so, the standard equation of the Hyperbola will be ;

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

By comparing the standard parameter (Vertices and eccentricity) with the given one, we get

$a=7$ and

$e=\frac{c}{a}=\frac{c}{7}=\frac{4}{3}$

From here,

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

Now, As we know the relation in a hyperbola

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

$b^2=c^2-a^2$

$b^2=\left(\frac{28}{3}\right)^2-7^2$

$b^2=\left(\frac{784}{9}\right)-49$

$b^2=\left(\frac{784-441}{9}\right)=\frac{343}{9}$

Hence, The Equation of the hyperbola is ;

$\frac{x^2}{49}-\frac{9y^2}{343}=1$

Question:15 Find the equations of the hyperbola satisfying the given conditions.

Foci $(0,\pm\sqrt{10})$, passing through (2,3)

Answer:

Given, in a hyperbola,

Foci $(0,\pm\sqrt{10})$, passing through (2,3)

Since foci of the hyperbola are in Y-axis, the equation of the hyperbola will be of the form ;

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

By comparing standard parameter (foci) with the given one, we get

$c=\sqrt{10}$

Now As we know, in a hyperbola

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

$a^2+b^2=10\:\:\:\:\:\:\:....(1)$

Now As the hyperbola passes through the point (2,3)

$\frac{3^2}{a^2}-\frac{2^2}{b^2}=1$

$9b^2-4a^2=a^2b^2\:\;\;\:\:\;\:....(2)$

Solving Equation (1) and (2)

$9(10-a^2)-4a^2=a^2(10-a^2)$

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

$(a^2)^2-18a^2-5a^2+90=0$

$(a^2-18)(a^2-5)=0$

$a^2=18\:or\:5$

Now, as we know that in a hyperbola $c$ is always greater than, $a$ we choose the value

$a^2=5$

$b^2=10-a^2=10-5=5$

Hence The Equation of the hyperbola is

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