# Class 11 RD Sharma Solutions – Chapter 27 Hyperbola – Exercise 27.1

**Question 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: Focus = (-1, 1) and Eccentricity = 3

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

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

We know, e = PF/PM ⇒ PF

^{2}= e^{2}PM^{2}⇒ 2{x

^{2}+ 1 + 2x + y^{2}+ 1 – 2y} = 9{x^{2}+ y^{2}+ 9 + 6x – 6y – 2xy}⇒ 2x

^{2}+ 2 + 4x + 2y^{2}+ 2 – 4y = 9x^{2}+ 9y^{2}+ 81 + 54x – 54y – 18xy⇒ 2x

^{2}+ 4 + 4x + 2y^{2}– 4y – 9x^{2}– 9y^{2}– 81 – 54x + 54y + 18xy = 0⇒ – 7x

^{2}– 7y^{2}– 50x + 50y + 18xy – 77 = 0⇒ 7(x

^{2}+ y^{2}) – 18xy + 50x – 50y + 77 = 0

∴The equation of hyperbola is 7(x^{2}+ y^{2}) – 18xy + 50x – 50y + 77 = 0.

**Question 2. Find the equation of the hyperbola whose**

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

**Solution:**

Given: Focus = (0, 3), Directrix => x + y – 1 = 0 and Eccentricity = 2

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

We know, e = PF/PM ⇒ PF

^{2}= e^{2}PM^{2}⇒ 2{x

^{2}+ y^{2}+ 9 – 6y} = 4{x^{2}+ y^{2}+ 1 – 2x – 2y + 2xy}⇒ 2x

^{2}+ 2y^{2}+ 18 – 12y – 4x^{2}– 4y^{2}– 4 – 8x + 8y – 8xy = 0⇒ – 2x

^{2 }– 2y^{2}– 8x – 4y – 8xy + 14 = 0⇒ –2(x

^{2}+ y^{2}– 4x + 2y + 4xy – 7) = 0⇒ x

^{2}+ y^{2}– 4x + 2y + 4xy – 7 = 0

∴The equation of hyperbola is x^{2}+ y^{2}– 4x + 2y + 4xy – 7 = 0.

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

**Solution:**

Focus = (1, 1), Directrix => 3x + 4y + 8 = 0 and Eccentricity = 2

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

We know, e = PF/PM ⇒ PF

^{2}= e^{2}PM^{2}⇒ 25{x

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

^{2}+ 25 – 50x + 25y^{2}+ 25 – 50y = 36x^{2}+ 64y^{2}+ 256 + 96xy + 256y + 192x⇒ 25x

^{2}+ 25 – 50x + 25y^{2}+ 25 – 50y – 36x^{2}– 64y^{2}– 256 – 96xy – 256y – 192x = 0⇒ – 11x

^{2}– 39y^{2}– 242x – 306y – 96xy – 206 = 0⇒ 11x

^{2}+ 96xy + 39y^{2}+ 242x + 306y + 206 = 0

∴The equation of hyperbola is 11x^{2}+ 96xy + 39y^{2}+ 242x + 306y + 206 = 0.

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

**Solution:**

Given: Focus = (1, 1), Directrix => 2x + y = 1 and Eccentricity =

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

We know, e = PF/PM ⇒ PF

^{2}= e^{2}PM^{2}⇒ 5{x

^{2}+ 1 – 2x + y^{2}+ 1 – 2y} = 3{4x^{2}+ y^{2}+ 1 + 4xy – 2y – 4x}⇒ 5x

^{2}+ 5 – 10x + 5y^{2}+ 5 – 10y = 12x^{2}+ 3y^{2}+ 3 + 12xy – 6y – 12x⇒ 5x

^{2}+ 5 – 10x + 5y^{2}+ 5 – 10y – 12x^{2}– 3y^{2}– 3 – 12xy + 6y + 12x = 0⇒ – 7x

^{2}+ 2y^{2}+ 2x – 4y – 12xy + 7 = 0⇒ 7x

^{2}+ 12xy – 2y^{2}– 2x + 4y– 7 = 0

∴The equation of hyperbola is 7x^{2}+ 12xy – 2y^{2 }– 2x + 4y– 7 = 0.

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

**Solution:**

Given: Focus = (2, -1), Directrix => 2x + 3y = 1 and Eccentricity = 2

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

^{2}= e^{2}PM^{2}⇒ 13{x

^{2 }+ 4 – 4x + y^{2}+ 1 + 2y} = 4{4x^{2}+ 9y^{2}+ 1 + 12xy – 6y – 4x}⇒ 13x

^{2}+ 52 – 52x + 13y^{2}+ 13 + 26y = 16x^{2}+ 36y^{2}+ 4 + 48xy – 24y – 16x⇒ 13x

^{2}+ 52 – 52x + 13y^{2}+ 13 + 26y – 16x^{2}– 36y^{2 }– 4 – 48xy + 24y + 16x = 0⇒ – 3x

^{2}– 23y^{2}– 36x + 50y – 48xy + 61 = 0⇒ 3x

^{2}+ 23y^{2}+ 48xy + 36x – 50y– 61 = 0

∴The equation of hyperbola is 3x^{2}+ 23y^{2}+ 48xy + 36x – 50y– 61 = 0.

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

**Solution:**

Given: Focus = (a, 0), Directrix => 2x + 3y = 1 and Eccentricity = 2

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

^{2}= e^{2}PM^{2}

⇒ 45{x

^{2}+ a^{2 }– 2ax + y2} = 16{4x^{2}+ y^{2 }+ a^{2}– 4xy – 2ay + 4ax}⇒ 45x

^{2}+ 45a^{2}– 90ax + 45y^{2}= 64x^{2}+ 16y^{2}+ 16a^{2}– 64xy – 32ay + 64ax⇒ 45x

^{2}+ 45a^{2}– 90ax + 45y^{2}– 64x^{2}– 16y^{2}– 16a^{2}+ 64xy + 32ay – 64ax = 0⇒ 19x

^{2}– 29y^{2}+ 154ax – 32ay – 64xy – 29a2 = 0

∴The equation of hyperbola is 19x^{2}– 29y^{2}+ 154ax – 32ay – 64xy – 29a^{2}= 0.

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

**Solution:**

Given: Focus = (2, 2), Directrix => x + y = 9 and Eccentricity = 2

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

^{2}= e^{2}PM^{2}⇒ x

^{2}+ 4 – 4x + y^{2}+ 4 – 4y = 2{x^{2}+ y^{2}+ 81 + 2xy – 18y – 18x}⇒ x

^{2}– 4x + y^{2}+ 8 – 4y = 2x^{2}+ 2y^{2}+ 162 + 4xy – 36y – 36x⇒ x

^{2}– 4x + y^{2}+ 8 – 4y – 2x^{2 }– 2y^{2}– 162 – 4xy + 36y + 36x = 0⇒ – x

^{2}– y^{2}+ 32x + 32y + 4xy – 154 = 0⇒ x

^{2}+ 4xy + y^{2}– 32x – 32y + 154 = 0

∴The equation of hyperbola is x^{2}+ 4xy + y^{2}– 32x – 32y + 154 = 0.

**Question 3. Find the eccentricity, coordinates of the foci, equations of directrices**,** and length of the latus-rectum of the hyperbola.**

**(i) 9x**^{2} – 16y^{2} = 144

^{2}– 16y

^{2}= 144

**Solution:**

Given: 9x

^{2}– 16y^{2}= 144This is of the form where, a

^{2}= 16, b^{2}= 9 i.e., a = 4 and b = 3Eccentricity is given by:

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

Foci = (±5, 0)

The equation of directrices is given as:⇒5x ∓ 16 = 0The length of latus-rectum is given as: 2b

^{2}/a = 2(9)/4

Length of latus rectum= 9/2

**(ii) 16x**^{2} – 9y^{2} = –144

^{2}– 9y

^{2}= –144

**Solution:**

Given: 16x

^{2}– 9y^{2}= –144This is of the form where, a

^{2}= 9, b^{2}= 16 i.e., a = 3 and b = 4.Eccentricity is given by:

Eccentricity =

Foci:The coordinates of the foci are (0, ±be)(0, ±be) = (0, ±4(5/4))

= (0, ±5).

The equation of directrices is given as: x =⇒ 5x ∓ 16 = 0.

The length of latus-rectum is given as: 2a^{2}/b = 2(9)/4 = 9/2.

**(iii) 4x**^{2} – 3y^{2} = 36

^{2}– 3y

^{2}= 36

**Solution:**

Given: 4x

^{2}– 3y^{2}= 36This is of the form where, a

^{2}= 9, b^{2}= 12 i.e., a = 3 and b = √12Eccentricity is given by:

Eccentricity =

Foci: The coordinates of the foci are (±ae, 0)= (±ae, 0)= (±√21, 0)

The length of latus-rectum is given as= 2b^{2}/a = 2(12)/3 = 24/3 = 8

**(iv) 3x**^{2} – y^{2} = 4

^{2}– y

^{2}= 4

**Solution:**

Given: 3x

^{2}– y^{2}= 4This is of the form where, and b = 2

Eccentricity is given by:

Eccentricity = 2

Foci:The coordinates of the foci are (±ae, 0)= (±ae, 0) = ±(2/√3)(2) = ±4/√3

(±ae, 0) = (±4/√3, 0)

The length of latus-rectum is given as:= 2b2/a = 2(4)/[2/√3] = 4√3.

**(v) 2x**^{2} – 3y^{2} = 5

^{2}– 3y

^{2}= 5

**Solution:**

Given: 2x

^{2}– 3y^{2}= 5This is of the form where, and

Eccentricity is given by:

Eccentricity =.

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

The length of latus-rectum is given as: 2b=^{2}/a

**Question 4. Find the axes, eccentricity, latus-rectum**,** and the coordinates of the foci of the hyperbola 25x**^{2} – 36y^{2} = 225.

^{2}– 36y

^{2}= 225.

**Solution:**

Given: 25x

^{2}– 36y^{2}= 225This 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) = (± √61/2, 0)

The length of latus-rectum is given as: 2b

^{2}/a

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

**Question 5. Find the **center**, eccentricity, foci**,** and directions of the hyperbola**

**(i) 16x**^{2} – 9y^{2} + 32x + 36y – 164 = 0

^{2}– 9y

^{2}+ 32x + 36y – 164 = 0

**Solution:**

Given: 16x

^{2}– 9y^{2}+ 32x + 36y – 164 = 0.⇒ 16x

^{2}+ 32x + 16 – 9y^{2}+ 36y – 36 – 16 + 36 – 164 = 0⇒ 16(x

^{2}+ 2x + 1) – 9(y^{2}– 4y + 4) – 16 + 36 – 164 = 0⇒ 16(x

^{2}+ 2x + 1) – 9(y^{2}– 4y + 4) – 144 = 0⇒ 16(x + 1)

^{2}– 9(y – 2)^{2}= 144Here, 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)

∴ 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) x**^{2} – y^{2} + 4x = 0

^{2}– y

^{2}+ 4x = 0

**Solution:**

Given: x

^{2}– y^{2}+ 4x = 0.⇒ x

^{2}– y^{2 }+ 4x = 0⇒ x

^{2}+ 4x + 4 – y^{2}– 4 = 0⇒ (x + 2)

^{2}– y^{2}= 4Here, center of the hyperbola is (2, 0).

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)

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

**(iii) x**^{2} – 3y^{2} – 2x = 8

^{2}– 3y

^{2}– 2x = 8

**Solution:**

Given: x

^{2}– 3y^{2}– 2x = 8.⇒ x

^{2}– 3y^{2}– 2x = 8⇒ x

^{2}– 2x + 1 – 3y^{2}– 1 = 8⇒ (x – 1)

^{2}– 3y^{2}= 9Here, center of the hyperbola is (1, 0)

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)

∴ The center is (1, 0), eccentricity (e) = 2√3/3, Foci = (1 ± 2√3, 0), Equation of directrix = X = 1±9/2√3.

**Question 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**

**Solution:**

Given: Distance between the foci = 16 and Eccentricity = √2

Let us compare with the equation of the form …..(1)

Distance between the foci is 2ae and b

^{2}= a^{2}(e^{2}– 1)So, 2ae = 16

⇒ ae = 16/2

⇒ a√2 = 8

⇒ a = 8/√2

⇒ a

^{2}= 64/2 = 32We know that, b

^{2}= a^{2}(e^{2}– 1)So, b

^{2}= 32[(√2)2 – 1]= 32(2 – 1)

= 32

The Equation of hyperbola is given as

⇒ x

^{2}– y^{2}= 32

∴ The Equation of hyperbola is x^{2}– y^{2}= 32.

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

**Solution:**

Given: Conjugate axis = 5 and Distance between foci = 13

Let us compare with the equation of the form …..(1)

Distance between the foci is 2ae and b

^{2}= a^{2}(e^{2}– 1)Length of conjugate axis is 2b

So, 2b = 5

⇒ b = 5/2

⇒ b

^{2}= 25/4We know that, 2ae = 13

ae = 13/2

⇒ a

^{2}e^{2}= 169/4b

^{2}= a^{2}(e^{2 }– 1)⇒ b

^{2}= a^{2}e^{2}– a^{2}⇒ 25/4 = 169/4 – a

^{2}⇒ a

^{2}= 169/4 – 25/4⇒ a

^{2}= 144/4 = 36The Equation of hyperbola is given as

∴ The Equation of hyperbola is 25x^{2}– 144y^{2}= 900.

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

**Solution:**

Given: Conjugate axis = 7 and Passes through the point (3, -2)

Conjugate axis is 2b

So, 2b = 7

⇒ b = 7/2

⇒ b

^{2}= 49/4The Equation of hyperbola is given as

Since it passes through points (3, -2), we have

⇒ a

^{2}= 441/65The equation of hyperbola is given as:

∴ The Equation of hyperbola is 65x^{2}– 36y^{2}= 441.

**Question 7. Find the equation of the hyperbola whose:**

**(i) foci are (6,4) and (-4,4) and eccentricity is 2.**

**Solution:**

Clearly, coordinates of the center are (1,4).

Equation of the hyperbola is:

Distance between the foci = 2ae

⇒ 2ae = 10

⇒ a = 5/2

⇒ a

^{2}= 25/4Since, b

^{2}= a^{2}(e^{2}– 1)⇒ b

^{2}= 75/4Putting the values in the equation, we get

⇒ 12×2 – 4y2 – 24x + 32y -127 = 0.

**(ii) vertices are (-8,-1) and (16,-1) and focus is (17,-1)**

**Solution:**

Clearly, coordinates of the center are (4,-1).

Equation of the hyperbola is :

Distance between vertices = 2ae

⇒ 24 = 2a

⇒ a = 12

⇒ a

^{2}= 144 and e^{2}= 169/144Since, b

^{2}= a^{2}(e^{2}– 1)⇒ b

^{2 }= 25Putting the values in the equation, we get

⇒ 25x^{2}– 144y^{2}– 200x – 288y – 3344 = 0.

**(iii) foci are (4,2) and (8,2) and eccentricity is 2.**

**Solution:**

Clearly, coordinates of the center are (6,2).

Equation of the hyperbola is :

Distance between the foci = 2ae

⇒ 2ae = 4

⇒ a = 1

Since, b

^{2}= a^{2}(e^{2}– 1)⇒ b

^{2}=3Putting the values in the equation, we get

⇒ 3×2 – y2 – 36x + 4y + 101 = 0.

**(iv) vertices are (0, ±7) and foci at (0, ±28/3).**

**Solution:**

Vertices of coordinates are (0, ±b) and (0, ±be).

⇒ b = 7

⇒ b

^{2}= 49and, be = 28/3

⇒ e = 4/3 ⇒ e

^{2}= 16/9Now, a

^{2}= b^{2}(e^{2}-1)⇒ a

^{2}= 343/9

The equation becomes:

**Question 8. Find the eccentricity if **the **length of **the **conjugate axis is 3/4 of the length of **the **traverse **axis**.**

**Solution:**

Given: 2b = 6a/4

⇒ b/a = 3/4

⇒ b

^{2}/a^{2}= 9/16Now,

e = 5/4.

**Question 9. Find the equation of the hyperbola whose focus is at (5,2) and (4,2) and **center** at (3,2).**

**Solution:**

Clearly the coordinates of the first vertex are (2,2).

Equation of the hyperbola is :

Distance between 2 vertices = 2a

⇒ a = 1

and, e = 2

b

^{2}= a^{2}(e^{2}– 1)⇒ b

^{2}= 3The equation becomes:

⇒ 3(x-3)^{2}– (y-2)^{2}= 3.

**Question 10. If P is any point on the hyperbola whose axis **is** equal, prove that SP.S’P = CP**^{2}.

^{2}.

**Solution:**

Given: a = b

Equation becomes: x

^{2}– y^{2}= a^{2}, C = (0,0), and

SP. S’P = 4a

^{4}+ 4a^{2}(a^{2 }+ b^{2}) + (a^{2}+ b^{2})^{2}– 8a^{2}b^{2}= (a

^{2}+ b^{2})^{2}= CP

Hence, SP.S’P = CP^{2}.

**Question 11. ** **Find the equation of the hyperbola whose:**

**(i) foci are (±2,0) and foci are (±3,0).**

**Solution:**

Equation of the hyperbola is :

Distance between the foci = 2ae

⇒ a = 2

⇒ a

^{2}= 4e = 3/2

Since, b

^{2}= a^{2}(e^{2}– 1)⇒ b

^{2}= 5

Putting the values in the equation, we get

**(ii) vertices are (0, ±4) and foci at (0, ±2/3).**

**Solution:**

Vertices of coordinates are (0, ±b) and (0, ±be).

⇒ b = 4

⇒ b

^{2}= 16and, be = 2/3

⇒ e = 2/3 ⇒ e

^{2}= 4/9Now, a

^{2}= b^{2}(e^{2}-1)⇒ a

^{2}= 343/9The equation becomes:

**Question 12. Find the equation when **the **distance between foci is 16 and eccentricity is** .

**Solution:**

Distance between foci = 2ae = 16

or, b

^{2}= 32

Equation becomes: x^{2}– y^{2}= 32.

**Question 13. Show that the set of all points such that the difference of their distance from (4,0) and (-4,0) is always equal to 2 represents a hyperbola.**

**Solution:**

Let P(x, y) be the point of the set.

Distance of P from (4,0) =

Distance of P from (-4,0) =

Given:

Squaring both sides, we have

⇒ 15x

^{2}– y^{2}= 15.

Thus, P represents a hyperbola.