RD Sharma Solutions for Class 11 Maths Chapter 27 Hyperbola

We have discussed in earlier chapters that a hyperbola is a particular case of conic. Now in this section, we shall study about hyperbola in detail by finding the equation of the hyperbola in standard form. Exercise-wise solutions are provided to help students understand the concepts clearly from the exam point of view. Experts have designed solutions where the concepts are explained in detail, which is very helpful in preparing for their board exams. Students who find difficulty in solving problems can quickly jump to RD Sharma Class 11 Maths Solutions, download the pdf from the links given below and can start practising offline for good results.

Chapter 27 – Hyperbola contains one exercise and the RD Sharma Solutions present in this page provide solutions to the questions present in this exercise. Now, let us have a look at the concepts discussed in this chapter.

  • Equation of the hyperbola in standard form.
    • Tracing of a hyperbola.
    • Second focus and second directrix of the hyperbola.
    • Various elements of a hyperbola.
    • Eccentricity.
    • Length of the latus-rectum.
    • Focal distances of a point.
    • Conjugate hyperbola.

Download the pdf of RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola

 

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Acccess RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola

EXERCISE 27.1 PAGE NO: 27.13

1. The equation of the directrix of a hyperbola is x – y + 3 = 0. Its focus is (-1, 1) and eccentricity 3. Find the equation of the hyperbola.

Solution:

Given:

The equation of the directrix of a hyperbola => x – y + 3 = 0.

Focus = (-1, 1) and

Eccentricity = 3

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any point of the hyperbola.

By using the formula,

e = PF/PM

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]

So,

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 1

[We know that (a – b)2 = a2 + b2 + 2ab &(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac]

So, 2{x2 + 1 + 2x + y2 + 1 – 2y} = 9{x2 + y2+ 9 + 6x – 6y – 2xy}

2x2 + 2 + 4x + 2y2 + 2 – 4y = 9x2 + 9y2+ 81 + 54x – 54y – 18xy

2x2 + 4 + 4x + 2y2– 4y – 9x2 – 9y2 – 81 – 54x + 54y + 18xy = 0

– 7x2 – 7y2 – 50x + 50y + 18xy – 77 = 0

7(x2 + y2) – 18xy + 50x – 50y + 77 = 0

∴The equation of hyperbola is 7(x2 + y2) – 18xy + 50x – 50y + 77 = 0

2. Find the equation of the hyperbola whose

(i) focus is (0, 3), directrix is x + y – 1 = 0 and eccentricity = 2

(ii) focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2

(iii) focus is (1, 1) directrix is 2x + y = 1 and eccentricity =√3

(iv) focus is (2, -1), directrix is 2x + 3y = 1 and eccentricity = 2

(v) focus is (a, 0), directrix is 2x + 3y = 1 and eccentricity = 2

(vi)focus is (2, 2), directrix is x + y = 9 and eccentricity = 2

Solution:

(i) focus is (0, 3), directrix is x + y – 1 = 0 and eccentricity = 2

Given:

Focus = (0, 3)

Directrix => x + y – 1 = 0

Eccentricity = 2

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any point of the hyperbola.

By using the formula,

e = PF/PM

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]

So,

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 2

[We know that (a – b)2 = a2 + b2 + 2ab &(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac]

So, 2{x2 + y2 + 9 – 6y} = 4{x2 + y2 + 1 – 2x – 2y + 2xy}

2x2 + 2y2 + 18 – 12y = 4x2 + 4y2+ 4 – 8x – 8y + 8xy

2x2 + 2y2 + 18 – 12y – 4x2 – 4y2 – 4 – 8x + 8y – 8xy = 0

– 2x2 – 2y2 – 8x – 4y – 8xy + 14 = 0

-2(x2 + y2 – 4x + 2y + 4xy – 7) = 0

x2 + y2 – 4x + 2y + 4xy – 7 = 0

∴The equation of hyperbola is x2 + y2 – 4x + 2y + 4xy – 7 = 0

(ii) focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2

Focus = (1, 1)

Directrix => 3x + 4y + 8 = 0

Eccentricity = 2

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any point of the hyperbola.

By using the formula,

e = PF/PM

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]

So,

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 3

[We know that (a – b)2 = a2 + b2 + 2ab &(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac]

25{x2 + 1 – 2x + y2 + 1 – 2y} = 4{9x2 + 16y2+ 64 + 24xy + 64y + 48x}

25x2 + 25 – 50x + 25y2 + 25 – 50y = 36x2 + 64y2 + 256 + 96xy + 256y + 192x

25x2 + 25 – 50x + 25y2 + 25 – 50y – 36x2 – 64y2 – 256 – 96xy – 256y – 192x = 0

– 11x2 – 39y2 – 242x – 306y – 96xy – 206 = 0

11x2 + 96xy + 39y2 + 242x + 306y + 206 = 0

∴The equation of hyperbola is11x2 + 96xy + 39y2 + 242x + 306y + 206 = 0

(iii) focus is (1, 1) directrix is 2x + y = 1 and eccentricity =√3

Given:

Focus = (1, 1)

Directrix => 2x + y = 1

Eccentricity =√3

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any point of the hyperbola.

By using the formula,

e = PF/PM

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]

So,

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 4

[We know that (a – b)2 = a2 + b2 + 2ab &(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac]

5{x2 + 1 – 2x + y2 + 1 – 2y} = 3{4x2 + y2+ 1 + 4xy – 2y – 4x}

5x2 + 5 – 10x + 5y2 + 5 – 10y = 12x2 + 3y2 + 3 + 12xy – 6y – 12x

5x2 + 5 – 10x + 5y2 + 5 – 10y – 12x2 – 3y2 – 3 – 12xy + 6y + 12x = 0

– 7x2 + 2y2 + 2x – 4y – 12xy + 7 = 0

7x2 + 12xy – 2y2 – 2x + 4y– 7 = 0

∴The equation of hyperbola is7x2 + 12xy – 2y2 – 2x + 4y– 7 = 0

(iv) focus is (2, -1), directrix is 2x + 3y = 1 and eccentricity = 2

Given:

Focus = (2, -1)

Directrix => 2x + 3y = 1

Eccentricity = 2

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any point of the hyperbola.

By using the formula,

e = PF/PM

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]

So,

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 5

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 6

[We know that (a – b)2 = a2 + b2 + 2ab &(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac]

13{x2 + 4 – 4x + y2 + 1 + 2y} = 4{4x2 + 9y2 + 1 + 12xy – 6y – 4x}

13x2 + 52 – 52x + 13y2 + 13 + 26y = 16x2 + 36y2 + 4 + 48xy – 24y – 16x

13x2 + 52 – 52x + 13y2 + 13 + 26y – 16x2 – 36y2 – 4 – 48xy + 24y + 16x = 0

– 3x2 – 23y2 – 36x + 50y – 48xy + 61 = 0

3x2 + 23y2 + 48xy + 36x – 50y– 61 = 0

∴The equation of hyperbola is3x2 + 23y2 + 48xy + 36x – 50y– 61 = 0

(v) focus is (a, 0), directrix is 2x + 3y = 1 and eccentricity = 2

Given:

Focus = (a, 0)

Directrix => 2x + 3y = 1

Eccentricity = 2

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any point of the hyperbola.

By using the formula,

e = PF/PM

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]

So,

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 7

[We know that (a – b)2 = a2 + b2 + 2ab &(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac]

45{x2 + a2 – 2ax + y2} = 16{4x2 + y2 + a2 – 4xy – 2ay + 4ax}

45x2 + 45a2 – 90ax + 45y2 = 64x2 + 16y2 + 16a2 – 64xy – 32ay + 64ax

45x2 + 45a2 – 90ax + 45y2 – 64x2 – 16y2 – 16a2 + 64xy + 32ay – 64ax = 0

19x2 – 29y2 + 154ax – 32ay – 64xy – 29a2 = 0

∴The equation of hyperbola is19x2 – 29y2 + 154ax – 32ay – 64xy – 29a2 = 0

(vi) focus is (2, 2), directrix is x + y = 9 and eccentricity = 2

Given:

Focus = (2, 2)

Directrix => x + y = 9

Eccentricity = 2

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any point of the hyperbola.

By using the formula,

e = PF/PM

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]

So,

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 8

[We know that (a – b)2 = a2 + b2 + 2ab &(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac]

x2 + 4 – 4x + y2 + 4 – 4y = 2{x2 + y2 + 81 + 2xy – 18y – 18x}

x2 – 4x + y2 + 8 – 4y = 2x2 + 2y2 + 162 + 4xy – 36y – 36x

x2 – 4x + y2 + 8 – 4y – 2x2 – 2y2 – 162 – 4xy + 36y + 36x = 0

– x2 – y2 + 32x + 32y + 4xy – 154 = 0

x2 + 4xy + y2 – 32x – 32y + 154 = 0

∴The equation of hyperbola isx2 + 4xy + y2 – 32x – 32y + 154 = 0

3. Find the eccentricity, coordinates of the foci, equations of directrices and length of the latus-rectum of the hyperbola.
(i) 9x2 – 16y2 = 144

(ii) 16x2 – 9y2 = -144

(iii) 4x2 – 3y2 = 36

(iv) 3x2 – y2 = 4

(v) 2x2 – 3y2 = 5

Solution:

(i) 9x2 – 16y2 = 144

Given:

The equation => 9x2 – 16y2 = 144

The equation can be expressed as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 9

The obtained equation is of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 10

Where, a2 = 16, b2 = 9 i.e., a = 4 and b = 3

Eccentricity is given by:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 11

Foci: The coordinates of the foci are (0, ±be)

(0, ±be) = (0, ±4(5/4))

= (0, ±5)

The equation of directrices is given as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 12

5x ∓ 16 = 0

The length of latus-rectum is given as:

2b2/a

= 2(9)/4

= 9/2

(ii) 16x2 – 9y2 = -144

Given:

The equation => 16x2 – 9y2 = -144

The equation can be expressed as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 13

The obtained equation is of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 14

Where, a2 = 9, b2 = 16 i.e., a = 3 and b = 4

Eccentricity is given by:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 15

Foci: The coordinates of the foci are (0, ±be)

(0, ±be) = (0, ±4(5/4))

= (0, ±5)

The equation of directrices is given as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 16

The length of latus-rectum is given as:

2a2/b

= 2(9)/4

= 9/2

(iii) 4x2 – 3y2 = 36

Given:

The equation => 4x2 – 3y2 = 36

The equation can be expressed as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 17

The obtained equation is of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 18

Where, a2 = 9, b2 = 12 i.e., a = 3 and b = √12

Eccentricity is given by:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 19

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 20

Foci: The coordinates of the foci are (±ae, 0)

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 21

(±ae, 0) = (±√21, 0)

The equation of directrices is given as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 22

The length of latus-rectum is given as:

2b2/a

= 2(12)/3

= 24/3

= 8

(iv) 3x2 – y2 = 4

Given:

The equation => 3x2 – y2 = 4

The equation can be expressed as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 23

The obtained equation is of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 24

Where, a = 2/√3 and b = 2

Eccentricity is given by:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 25

Foci: The coordinates of the foci are (±ae, 0)

(±ae, 0) = ±(2/√3)(2) = ±4/√3

(±ae, 0) = (±4/√3, 0)

The equation of directrices is given as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 26

The length of latus-rectum is given as:

2b2/a

= 2(4)/[2/√3]

= 4√3

(v) 2x2 – 3y2 = 5

Given:

The equation => 2x2 – 3y2 = 5

The equation can be expressed as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 27

The obtained equation is of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 28

Where, a = √5/√2 and b = √5/√3

Eccentricity is given by:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 29

Foci: The coordinates of the foci are (±ae, 0)

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 30

(±ae, 0) = (±5/√6, 0)

The equation of directrices is given as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 31

The length of latus-rectum is given as:

2b2/a

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 32

4. Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola 25x2 – 36y2 = 225.

Solution:

Given:

The equation=> 25x2 – 36y2 = 225

The equation can be expressed as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 33

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 34

The obtained equation is of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 35

Where, a = 3 and b = 5/2

Eccentricity is given by:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 36

Foci: The coordinates of the foci are (±ae, 0)

(±ae, 0) = ±3 (√61/6) = ± √61/2

(±ae, 0) = (± √61/2, 0)

The equation of directrices is given as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 37

The length of latus-rectum is given as:

2b2/a

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 38

∴ Transverse axis = 6, conjugate axis = 5, e = √61/6, LR = 25/6, foci = (± √61/2, 0)

5. Find the centre, eccentricity, foci and directions of the hyperbola
(i) 16x2 – 9y2 + 32x + 36y – 164 = 0

(ii) x2 – y2 + 4x = 0

(iii) x2 – 3y2 – 2x = 8

Solution:

(i) 16x2 – 9y2 + 32x + 36y – 164 = 0

Given:

The equation => 16x2 – 9y2 + 32x + 36y – 164 = 0

Let us find the centre, eccentricity, foci and directions of the hyperbola

By using the given equation

16x2 – 9y2 + 32x + 36y – 164 = 0

16x2 + 32x + 16 – 9y2 + 36y – 36 – 16 + 36 – 164 = 0

16(x2 + 2x + 1) – 9(y2 – 4y + 4) – 16 + 36 – 164 = 0

16(x2 + 2x + 1) – 9(y2 – 4y + 4) – 144 = 0

16(x + 1)2 – 9(y – 2)2 = 144

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 39

Here, center of the hyperbola is (-1, 2)

So, let x + 1 = X and y – 2 = Y

The obtained equation is of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 40

Where, a = 3 and b = 4

Eccentricity is given by:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 41

Foci: The coordinates of the foci are (±ae, 0)

X = ±5 and Y = 0

x + 1 = ±5 and y – 2 = 0

x = ±5 – 1 and y = 2

x = 4, -6 and y = 2

So, Foci: (4, 2) (-6, 2)

Equation of directrix are:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 42

∴ The center is (-1, 2), eccentricity (e) = 5/3, Foci = (4, 2) (-6, 2), Equation of directrix = 5x – 4 = 0 and 5x + 14 = 0

(ii) x2 – y2 + 4x = 0

Given:

The equation => x2 – y2 + 4x = 0

Let us find the centre, eccentricity, foci and directions of the hyperbola

By using the given equation

x2 – y2 + 4x = 0

x2 + 4x + 4 – y2 – 4 = 0

(x + 2)2 – y2 = 4

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 43

Here, center of the hyperbola is (2, 0)

So, let x – 2 = X

The obtained equation is of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 44

Where, a = 2 and b = 2

Eccentricity is given by:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 45

Foci: The coordinates of the foci are (±ae, 0)

X = ± 2√2 and Y = 0

X + 2 = ± 2√2 and Y = 0

X= ± 2√2 – 2 and Y = 0

So, Foci = (± 2√2 – 2, 0)

Equation of directrix are:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 46

∴ The center is (-2, 0), eccentricity (e) = √2, Foci = (-2± 2√2, 0), Equation of directrix =

x + 2 = ±√2

(iii) x2 – 3y2 – 2x = 8

Given:

The equation => x2 – 3y2 – 2x = 8

Let us find the centre, eccentricity, foci and directions of the hyperbola

By using the given equation

x2 – 3y2 – 2x = 8

x2 – 2x + 1 – 3y2 – 1 = 8

(x – 1)2 – 3y2 = 9

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 47

Here, center of the hyperbola is (1, 0)

So, let x – 1 = X

The obtained equation is of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 48

Where, a = 3 and b = √3

Eccentricity is given by:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 49

Foci: The coordinates of the foci are (±ae, 0)

X = ± 2√3 and Y = 0

X – 1 = ± 2√3 and Y = 0

X= ± 2√3 + 1 and Y = 0

So, Foci = (1 ± 2√3, 0)

Equation of directrix are:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 50

∴ The center is (1, 0), eccentricity (e) = 2√3/3, Foci = (1 ± 2√3, 0), Equation of directrix =

X = 1±9/2√3

6. Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases:
(i) the distance between the foci = 16 and eccentricity = √2

(ii) conjugate axis is 5 and the distance between foci = 13

(iii) conjugate axis is 7 and passes through the point (3, -2)

Solution:

(i) the distance between the foci = 16 and eccentricity = √2

Given:

Distance between the foci = 16

Eccentricity = √2

Let us compare with the equation of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 51

Distance between the foci is 2ae and b2 = a2(e2 – 1)

So,

2ae = 16

ae = 16/2

a√2 = 8

a = 8/√2

a2 = 64/2

= 32

We know that, b2 = a2(e2 – 1)

So, b2 = 32 [(√2)2 – 1]

= 32 (2 – 1)

= 32

The Equation of hyperbola is given as

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 52

x2 – y2 = 32

∴ The Equation of hyperbola is x2 – y2 = 32

(ii) conjugate axis is 5 and the distance between foci = 13

Given:

Conjugate axis = 5

Distance between foci = 13

Let us compare with the equation of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 53

Distance between the foci is 2ae and b2 = a2(e2 – 1)

Length of conjugate axis is 2b

So,

2b = 5

b = 5/2

b2 = 25/4

We know that, 2ae = 13

ae = 13/2

a2e2 = 169/4

b2 = a2(e2 – 1)

b2 = a2e2 – a2

25/4 = 169/4 – a2

a2 = 169/4 – 25/4

= 144/4

= 36

The Equation of hyperbola is given as

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 54

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 55

∴ The Equation of hyperbola is 25x2 – 144y2 = 900

(iii) conjugate axis is 7 and passes through the point (3, -2)

Given:

Conjugate axis = 7

Passes through the point (3, -2)

Conjugate axis is 2b

So,

2b = 7

b = 7/2

b2 = 49/4

The Equation of hyperbola is given as

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 56

Since it passes through points (3, -2)

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 57

a2 = 441/65

The equation of hyperbola is given as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 58

∴ The Equation of hyperbola is 65x2 – 36y2 = 441

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