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

Class 11: Maths Chapter 27 solutions. Complete Class 11 Maths Chapter 27 Notes.

RD Sharma Solutions for Class 11 Maths Chapter 27–Hyperbola

RD Sharma 11th Maths Chapter 27, Class 11 Maths Chapter 27 solutions

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,

[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,

[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,

[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,

[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,

[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,

[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,

[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:

The obtained equation is of the form

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

Eccentricity is given by:

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

(0, ±be) = (0, ±4(5/4))

= (0, ±5)

The equation of directrices is given as:

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:

The obtained equation is of the form

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

Eccentricity is given by:

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

(0, ±be) = (0, ±4(5/4))

= (0, ±5)

The equation of directrices is given as:

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:

The obtained equation is of the form

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

Eccentricity is given by:

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

(±ae, 0) = (±√21, 0)

The equation of directrices is given as:

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:

The obtained equation is of the form

Where, a = 2/√3 and b = 2

Eccentricity is given by:

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:

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:

The obtained equation is of the form

Where, a = √5/√2 and b = √5/√3

Eccentricity is given by:

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

(±ae, 0) = (±5/√6, 0)

The equation of directrices is given as:

The length of latus-rectum is given as:

2b2/a

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:

The obtained equation is of the form

Where, a = 3 and b = 5/2

Eccentricity is given by:

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:

The length of latus-rectum is given as:

2b2/a

∴ 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

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

So, let x + 1 = X and y – 2 = Y

The obtained equation is of the form

Where, a = 3 and b = 4

Eccentricity is given by:

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:

∴ 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

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

So, let x – 2 = X

The obtained equation is of the form

Where, a = 2 and b = 2

Eccentricity is given by:

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:

∴ 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

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

So, let x – 1 = X

The obtained equation is of the form

Where, a = 3 and b = √3

Eccentricity is given by:

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:

∴ 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

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

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

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

∴ 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

Since it passes through points (3, -2)

a2 = 441/65

The equation of hyperbola is given as:

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

RD Sharma Solutions for Class 11 Maths Chapter 27: Download PDF

RD Sharma Solutions for Class 11 Maths Chapter 27–Hyperbola

Download PDF: RD Sharma Solutions for Class 11 Maths Chapter 27–Hyperbola PDF

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About RD Sharma

RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.

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