# RD Sharma Solutions for Class 11 Maths Chapter 27 Hyperbola

## RD Sharma Solutions Class 11 Maths Chapter 27 â€“ Download Free PDF Updated for (2021-22)

RD Sharma Solutions for Class 11 Maths Chapter 27 – HyperbolaÂ are provided here for students to score good marks in the board exams. 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 RD Sharma 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

### Access 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 â€“ 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 â€“ y + 3 = 0.

Focus = (-1, 1) and

Eccentricity = 3

Now, let us find the equation of the hyperbola

Let â€˜Mâ€™ 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 thatÂ (a â€“ b)2Â = a2Â + b2Â + 2ab &(a + b + c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac]

So, 2{x2Â + 1 + 2x + y2Â + 1 â€“ 2y} = 9{x2Â + y2+ 9 + 6x â€“ 6y â€“ 2xy}

2x2Â + 2 + 4x + 2y2Â + 2 â€“ 4y = 9x2Â + 9y2+ 81 + 54x â€“ 54y â€“ 18xy

2x2Â + 4 + 4x + 2y2â€“ 4y â€“ 9x2Â – 9y2Â – 81 â€“ 54x + 54y + 18xy = 0

â€“ 7x2Â – 7y2Â â€“ 50x + 50y + 18xy â€“ 77 = 0

7(x2Â + y2) â€“ 18xy + 50x â€“ 50y + 77 = 0

âˆ´The equation of hyperbola is 7(x2Â + y2) â€“ 18xy + 50x â€“ 50y + 77 = 0

2. Find the equation of the hyperbola whose

(i) focus is (0, 3), directrix is x + y â€“ 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 =âˆš3

(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 â€“ 1 = 0 and eccentricity = 2

Given:

Focus = (0, 3)

Directrix => x + y â€“ 1 = 0

Eccentricity = 2

Now, let us find the equation of the hyperbola

Let â€˜Mâ€™ 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 â€“ b)2Â = a2Â + b2Â + 2ab &(a + b + c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac]

So, 2{x2Â + y2Â + 9 â€“ 6y} = 4{x2Â + y2Â + 1 â€“ 2x â€“ 2y + 2xy}

2x2Â + 2y2Â + 18 â€“ 12y = 4x2Â + 4y2+ 4 â€“ 8x â€“ 8y + 8xy

2x2Â + 2y2Â + 18 â€“ 12y â€“ 4x2Â â€“ 4y2Â â€“ 4 â€“ 8x + 8y â€“ 8xy = 0

â€“ 2x2Â â€“ 2y2Â â€“ 8x â€“ 4y â€“ 8xy + 14 = 0

-2(x2Â + y2Â â€“ 4x + 2y + 4xy â€“ 7) = 0

x2Â + y2Â â€“ 4x + 2y + 4xy â€“ 7 = 0

âˆ´The equation of hyperbola is x2Â + y2Â â€“ 4x + 2y + 4xy â€“ 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 â€˜Mâ€™ 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 â€“ b)2Â = a2Â + b2Â + 2ab &(a + b + c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac]

25{x2Â + 1 â€“ 2x + y2Â + 1 â€“ 2y} = 4{9x2Â + 16y2+ 64 + 24xy + 64y + 48x}

25x2Â + 25 â€“ 50x + 25y2Â + 25 â€“ 50y = 36x2Â + 64y2Â + 256 + 96xy + 256y + 192x

25x2Â + 25 â€“ 50x + 25y2Â + 25 â€“ 50y â€“ 36x2Â â€“ 64y2Â â€“ 256 â€“ 96xy â€“ 256y â€“ 192xÂ = 0

â€“ 11x2Â â€“ 39y2Â â€“ 242x â€“ 306y â€“ 96xy â€“ 206 = 0

11x2Â + 96xy + 39y2Â + 242x + 306y + 206 = 0

âˆ´The equation of hyperbola is11x2Â + 96xy + 39y2Â + 242x + 306y + 206 = 0

(iii) focus is (1, 1) directrix is 2x + y = 1 and eccentricity =âˆš3

Given:

Focus = (1, 1)

Directrix => 2x + y = 1

Eccentricity =âˆš3

Now, let us find the equation of the hyperbola

Let â€˜Mâ€™ 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 â€“ b)2Â = a2Â + b2Â + 2ab &(a + b + c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac]

5{x2Â + 1 â€“ 2x + y2Â + 1 â€“ 2y} = 3{4x2Â + y2+ 1 + 4xy â€“ 2y â€“ 4x}

5x2Â + 5 â€“ 10x + 5y2Â + 5 â€“ 10y = 12x2Â + 3y2Â + 3 + 12xy â€“ 6y â€“ 12x

5x2Â + 5 â€“ 10x + 5y2Â + 5 â€“ 10y â€“ 12x2Â â€“ 3y2Â â€“ 3 â€“ 12xy + 6y + 12xÂ = 0

â€“ 7x2Â + 2y2Â + 2x â€“ 4y â€“ 12xy + 7 = 0

7x2Â + 12xy â€“ 2y2Â â€“ 2x + 4yâ€“ 7 = 0

âˆ´The equation of hyperbola is7x2Â + 12xy â€“ 2y2Â â€“ 2x + 4yâ€“ 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 â€˜Mâ€™ 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 â€“ b)2Â = a2Â + b2Â + 2ab &(a + b + c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac]

13{x2Â + 4 â€“ 4x + y2Â + 1 + 2y} = 4{4x2Â + 9y2Â + 1 + 12xy â€“ 6y â€“ 4x}

13x2Â + 52 â€“ 52x + 13y2Â + 13 + 26y = 16x2Â + 36y2Â + 4 + 48xy â€“ 24y â€“ 16x

13x2Â + 52 â€“ 52x + 13y2Â + 13 + 26y â€“ 16x2Â â€“ 36y2Â â€“ 4 â€“ 48xy + 24y + 16xÂ = 0

â€“ 3x2Â â€“ 23y2Â â€“ 36x + 50y â€“ 48xy + 61 = 0

3x2Â + 23y2Â + 48xy + 36x â€“ 50yâ€“ 61 = 0

âˆ´The equation of hyperbola is3x2Â + 23y2Â + 48xy + 36x â€“ 50yâ€“ 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 â€˜Mâ€™ 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 â€“ b)2Â = a2Â + b2Â + 2ab &(a + b + c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac]

45{x2Â + a2Â â€“ 2ax + y2} = 16{4x2Â + y2Â + a2Â â€“ 4xy â€“ 2ay + 4ax}

45x2Â + 45a2Â â€“ 90ax + 45y2Â = 64x2Â + 16y2Â + 16a2Â â€“ 64xy â€“ 32ay + 64ax

45x2Â + 45a2Â â€“ 90ax + 45y2Â â€“ 64x2Â â€“ 16y2Â â€“ 16a2Â + 64xy + 32ay â€“ 64axÂ = 0

19x2Â â€“ 29y2Â + 154ax – 32ay – 64xy – 29a2Â = 0

âˆ´The equation of hyperbola is19x2Â â€“ 29y2Â + 154ax – 32ay – 64xy – 29a2Â = 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 â€˜Mâ€™ 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 â€“ b)2Â = a2Â + b2Â + 2ab &(a + b + c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac]

x2Â + 4 â€“ 4x + y2Â + 4 â€“ 4y = 2{x2Â + y2Â + 81 + 2xy â€“ 18y â€“ 18x}

x2Â â€“ 4x + y2Â + 8 â€“ 4y = 2x2Â + 2y2Â + 162 + 4xy â€“ 36y â€“ 36x

x2Â â€“ 4x + y2Â + 8 â€“ 4y â€“ 2x2Â â€“ 2y2Â â€“ 162 â€“ 4xy + 36y + 36xÂ = 0

â€“ x2Â â€“ y2Â + 32x + 32y + 4xy â€“ 154 = 0

x2Â + 4xy + y2Â â€“ 32x â€“ 32y + 154 = 0

âˆ´The equation of hyperbola isx2Â + 4xy + y2Â â€“ 32x â€“ 32y + 154 = 0

3. Find the eccentricity, coordinates of the foci, equations of directrices and length of the latus-rectum of the hyperbola.
(i) 9x2Â â€“ 16y2Â = 144

(ii) 16x2Â â€“ 9y2Â = -144

(iii) 4x2Â â€“ 3y2Â = 36

(iv) 3x2Â â€“ y2Â = 4

(v) 2x2Â â€“ 3y2Â = 5

Solution:

(i) 9x2Â â€“ 16y2Â = 144

Given:

The equation => 9x2Â â€“ 16y2Â = 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, Â±be)

(0, Â±be) = (0, Â±4(5/4))

= (0, Â±5)

The equation of directrices is given as:

5x âˆ“ 16 = 0

The length of latus-rectum is given as:

2b2/a

= 2(9)/4

= 9/2

(ii) 16x2Â â€“ 9y2Â = -144

Given:

The equation => 16x2Â â€“ 9y2Â = -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, Â±be)

(0, Â±be) = (0, Â±4(5/4))

= (0, Â±5)

The equation of directrices is given as:

The length of latus-rectum is given as:

2a2/b

= 2(9)/4

= 9/2

(iii) 4x2Â â€“ 3y2Â = 36

Given:

The equation => 4x2Â â€“ 3y2Â = 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 = âˆš12

Eccentricity is given by:

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

(Â±ae, 0) = (Â±âˆš21, 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) 3x2Â â€“ y2Â = 4

Given:

The equation => 3x2Â â€“ y2Â = 4

The equation can be expressed as:

The obtained equation is of the form

Where, a = 2/âˆš3 and b = 2

Eccentricity is given by:

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

(Â±ae, 0) = Â±(2/âˆš3)(2) = Â±4/âˆš3

(Â±ae, 0) = (Â±4/âˆš3, 0)

The equation of directrices is given as:

The length of latus-rectum is given as:

2b2/a

= 2(4)/[2/âˆš3]

= 4âˆš3

(v) 2x2Â â€“ 3y2Â = 5

Given:

The equation => 2x2Â â€“ 3y2Â = 5

The equation can be expressed as:

The obtained equation is of the form

Where, a = âˆš5/âˆš2 and b = âˆš5/âˆš3

Eccentricity is given by:

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

(Â±ae, 0) = (Â±5/âˆš6, 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 25x2Â â€“ 36y2Â = 225.

Solution:

Given:

The equation=> 25x2Â â€“ 36y2Â = 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 (Â±ae, 0)

(Â±ae, 0) = Â±3 (âˆš61/6) = Â± âˆš61/2

(Â±ae, 0) = (Â± âˆš61/2, 0)

The equation of directrices is given as:

The length of latus-rectum is given as:

2b2/a

âˆ´ Transverse axis = 6, conjugate axis = 5, e = âˆš61/6, LR = 25/6, foci = (Â± âˆš61/2, 0)

5. Find the centre, eccentricity, foci and directions of the hyperbola
(i) 16x2Â â€“ 9y2Â + 32x + 36y â€“ 164 = 0

(ii) x2Â â€“ y2Â + 4x = 0

(iii) x2Â â€“ 3y2Â â€“ 2x = 8

Solution:

(i) 16x2Â â€“ 9y2Â + 32x + 36y â€“ 164 = 0

Given:

The equation => 16x2Â â€“ 9y2Â + 32x + 36y â€“ 164 = 0

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

By using the given equation

16x2Â â€“ 9y2Â + 32x + 36y â€“ 164 = 0

16x2Â + 32x + 16 â€“ 9y2Â + 36y â€“ 36 â€“ 16 + 36 â€“ 164 = 0

16(x2Â + 2x + 1) â€“ 9(y2Â â€“ 4y + 4) â€“ 16 + 36 â€“ 164 = 0

16(x2Â + 2x + 1) â€“ 9(y2Â â€“ 4y + 4) â€“ 144 = 0

16(x + 1)2Â â€“ 9(y â€“ 2)2Â = 144

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

So, let x + 1 = X and y â€“ 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 (Â±ae, 0)

X =Â Â±5 and Y = 0

x + 1 = Â±5 and y â€“ 2 = 0

x = Â±5 â€“ 1 and y = 2

x = 4, -6 and y = 2

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

Equation of directrix are:

âˆ´ The center is (-1, 2), eccentricity (e) = 5/3, Foci = (4, 2) (-6, 2), Equation of directrix = 5x â€“ 4 = 0 and 5x + 14 = 0

(ii) x2Â â€“ y2Â + 4x = 0

Given:

The equation => x2Â â€“ y2Â + 4x = 0

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

By using the given equation

x2Â â€“ y2Â + 4x = 0

x2Â + 4x + 4 â€“ y2Â â€“ 4 = 0

(x + 2)2Â â€“ y2Â = 4

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

So, let x â€“ 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 (Â±ae, 0)

X = Â± 2âˆš2 and Y = 0

X + 2 = Â± 2âˆš2 and Y = 0

X= Â± 2âˆš2 â€“ 2 and Y = 0

So, Foci = (Â± 2âˆš2 â€“ 2, 0)

Equation of directrix are:

âˆ´ The center is (-2, 0), eccentricity (e) = âˆš2, Foci = (-2Â± 2âˆš2, 0), Equation of directrix =

x + 2 = Â±âˆš2

(iii) x2Â â€“ 3y2Â â€“ 2x = 8

Given:

The equation => x2Â â€“ 3y2Â â€“ 2x = 8

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

By using the given equation

x2Â â€“ 3y2Â â€“ 2x = 8

x2Â â€“ 2x + 1 â€“ 3y2Â â€“ 1 = 8

(x â€“ 1)2Â â€“ 3y2Â = 9

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

So, let x â€“ 1 = X

The obtained equation is of the form

Where, a = 3 and b = âˆš3

Eccentricity is given by:

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

X = Â± 2âˆš3 and Y = 0

X â€“ 1 = Â± 2âˆš3 and Y = 0

X= Â± 2âˆš3 + 1 and Y = 0

So, Foci = (1 Â± 2âˆš3, 0)

Equation of directrix are:

âˆ´ The center is (1, 0), eccentricity (e) = 2âˆš3/3, Foci = (1 Â± 2âˆš3, 0), Equation of directrix =

X = 1Â±9/2âˆš3

6. Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases:
(i) the distanceÂ between the foci = 16 and eccentricity = âˆš2

(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 distanceÂ between the foci = 16 and eccentricity = âˆš2

Given:

DistanceÂ between the foci = 16

Eccentricity = âˆš2

Let us compare with the equation of the form

Distance between the foci is 2ae and b2Â = a2(e2Â â€“ 1)

So,

2ae = 16

ae = 16/2

aâˆš2 = 8

a = 8/âˆš2

a2 = 64/2

= 32

We know that, b2Â = a2(e2Â â€“ 1)

So, b2 = 32 [(âˆš2)2 â€“ 1]

= 32 (2 – 1)

= 32

The Equation of hyperbola is given as

x2 â€“ y2 = 32

âˆ´ The Equation of hyperbola is x2 â€“ 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 b2Â = a2(e2Â â€“ 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

b2Â = a2(e2Â â€“ 1)

b2Â = a2e2Â â€“ a2

25/4 = 169/4 â€“ a2

a2 = 169/4 â€“ 25/4

= 144/4

= 36

The Equation of hyperbola is given as

âˆ´ The Equation of hyperbola is 25x2 â€“ 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:

âˆ´ The Equation of hyperbola is 65x2 â€“ 36y2 = 441

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