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

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

Edited By Ravindra Pindel | Updated on Jul 18, 2022 11:54 AM IST

In the previous exercises of this chapter, you have already learned about circles, parabola, and ellipse. In the NCERT solutions for Class 11 Maths chapter 11 exercise 11.4, you will learn about another conic section called 'Hyperbola'. The set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant is called a hyperbola. In the real world like open orbits of some comets about the sun follow hyperbolas. Hyperbola has applications in the field of science, research, and design in the design of bridges.

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The eccentricity, standard equation of hyperbola, latus rectum are also covered in the Class 11 Maths chapter 11 exercise 11.4 solution. You are advised to go through the theories, definitions, and observations given in the NCERT textbook before exercise 11.4 Class 11 Maths. After that, you can try to solve the given solved examples and NCERT syllabus problems by yourself. You can Class 11 Maths chapter 11 exercise 11.4 solutions whenever find difficulty while solving them. Click on the NCERT Solutions link if you are looking for NCERT solutions at one place.

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## Conic Section Class 11 Chapter 11 Exercise 11.4

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

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 :

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

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

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 :

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

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

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 :

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

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

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 :

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

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

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 :

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

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

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 :

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

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 ;

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

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 ;

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

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 ;

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

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 ;

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

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 ;

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

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 ;

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

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 ;

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

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 ;

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

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

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

## More About NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.4:-

Class 11 Maths chapter 11 exercise 11.4 consists of questions related to finding the coordinates of the foci of the hyperbola, vertices of the hyperbola, the eccentricity of the hyperbola, and the length of the latus rectum of the hyperbola. Also, you will get questions related to writing the equation of hyperbola where the other parameters of the hyperbola are given.

Also Read| Conic Section Class 11 Notes

## Benefits of NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.4:-

• NCERT solutions for Class 11 Maths chapter 11 exercise 11.4 are designed by subject matter experts who are experienced in this field.
• Class 11 Maths chapter 11 exercise 11.4 solutions are explained in a very detailed manner which could be understood by an average student also.
• You can use these Class 11 Maths chapter 11 exercise 11.4 solutions to get help in your homework.

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### Frequently Asked Questions (FAQs)

1. What is the definition of hyperbola ?

The set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant is called a hyperbola.

2. What is the centre of the hyperbola ?

The center of the hyperbola is the mid-point of the line segment joining the foci.

3. What is the transverse axis of the hyperbola ?

The transverse axis is the line passing through the foci of the hyperbola.

4. What is the conjugate axis of the hyperbola ?

The conjugate axis is called the line passing through the centre of the hyperbola and perpendicular to the transverse axis.

5. Can the eccentricity of hyperbola be less than one ?

No, the eccentricity of a hyperbola is always greater than one.

6. what is the eccentricity of parabola ?

The eccentricity of a parabola is always one.

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