RD Sharma Solutions for Class 11 Maths Chapter 27 Hyperbola

RD Sharma Solutions Class 11 Maths Chapter 27 – Download Free PDF Updated for (2023-24)

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola are provided here for students to enhance skills in Mathematics and 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.

Chapter 27 – Hyperbola contains one exercise and the RD Sharma Solutions provides answers in a descriptive manner to the questions present in this exercise. Students who find difficulty in solving problems can quickly jump to  RD Sharma Class 11 Maths Solutions in PDF format from the links given below and can start practising offline for good results. 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.

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola

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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 the 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 the hyperbola to the directrix]

So,

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 the 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 the hyperbola to the directrix]

So,

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 the 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 the hyperbola to the directrix]

So,

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 the 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 the hyperbola to the directrix]

So,

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 the 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 the hyperbola to the directrix]

So,

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 the 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 the hyperbola to the directrix]

So,

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 the 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 the hyperbola to the directrix]

So,

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 the 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 the 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 the 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 the 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 the 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 the 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, centre 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 is:

∴ The centre 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, centre 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 is:

∴ The centre 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, centre 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 is:

∴ The centre 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)

The length of the 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)

The 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 the hyperbola is given as:

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