Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 20 May, 2021

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 ⇒ PF2 = e2PM2



⇒ (x+1)^2 + (y-1)^2 = 3^2[\frac{x-y+3}{\sqrt{1^2+(-1)^2}}]

⇒ 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.

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 ⇒ PF2 = e2PM2 

⇒ (x-0)^2 + (y-3)^2 = 2^2[\frac{x+y-1}{\sqrt{1^2+(1)^2}}]^2

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

⇒ 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

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 ⇒ PF2 = e2PM2

⇒ (x-1)^2 + (y-1)^2 = 2^2[\frac{3x+4y+8}{\sqrt{3^2+(4)^2}}]^2

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

(iii) focus is (1, 1) directrix is 2x + y = 1 and eccentricity = \sqrt{3}

Solution:

Given: Focus = (1, 1), Directrix => 2x + y = 1 and Eccentricity = \sqrt{3}

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

We know, e = PF/PM ⇒ PF2 = e2PM2 

⇒ (x-1)^2 + (y-1)^2 = (\sqrt{3})^2[\frac{2x+y-1}{\sqrt{2^2+(1)^2}}]^2

⇒ 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 is 7x2 + 12xy – 2y2 – 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 ⇒ PF2 = e2PM2

⇒ (x-2)^2 + (y+1)^2 = 2^2[\frac{2x+3y-1}{\sqrt{2^2+(3)^2}}]^2

⇒ 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 is 3x2 + 23y2 + 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 ⇒ PF2 = e2PM2 

⇒ (x-a)^2 + (y-0)^2 = (4/3)^2[\frac{2x-y+a}{\sqrt{2^2+(-1)^2}}]^2     

⇒ 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 is 19x2 – 29y2 + 154ax – 32ay – 64xy – 29a2 = 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 ⇒ PF2 = e2PM2

⇒ (x-2)^2 + (y-2)^2 = (2)^2[\frac{x+y-9}{\sqrt{1^2+(1)^2}}]^2

⇒ 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 is x2 + 4xy + y2 – 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) 9x2 – 16y2 = 144

Solution:

Given: 9x2 – 16y2 = 144

⇒ \frac{9x^2}{144}-\frac{16y^2}{144} = 1

⇒ \frac{x^2}{16}-\frac{y^2}{9} = 1

This is of the form \frac{x^2}{a^2}-\frac{y^2}{b^2} = 1   where, a2 = 16, b2 = 9 i.e., a = 4 and b = 3

Eccentricity is given by:



e = \sqrt{1+\frac{b^2}{a^2}}

= \sqrt{1+\frac{9}{16}}

= \sqrt{\frac{25}{16}}

Eccentricity = \frac{5}{4}

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

Foci = (±5, 0)

The equation of directrices is given as: x = ±\frac{a}{e}   5x ∓ 16 = 0

The length of latus-rectum is given as: 2b2/a = 2(9)/4

Length of latus rectum= 9/2

(ii) 16x2 – 9y2 = –144

Solution:

Given: 16x2 – 9y2 = –144

⇒ \frac{16x^2}{144}-\frac{9y^2}{144} = -1

⇒ \frac{x^2}{9}-\frac{y^2}{16} = -1

This is of the form \frac{x^2}{a^2}-\frac{y^2}{b^2} = -1   where, a2 = 9, b2 = 16 i.e., a = 3 and b = 4.

Eccentricity is given by:

e = \sqrt{1+\frac{a^2}{b^2}}

= \sqrt{1+\frac{9}{16}}

= \sqrt{\frac{25}{16}}

Eccentricity = \frac{5}{4}

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 = ±\frac{b}{e} ⇒ 5x ∓ 16 = 0.

The length of latus-rectum is given as: 2a2/b = 2(9)/4 = 9/2.

(iii) 4x2 – 3y2 = 36

Solution:

Given: 4x2 – 3y2 = 36

⇒ \frac{4x^2}{36}-\frac{3y^2}{36} = 1

⇒ \frac{x^2}{9}-\frac{y^2}{12} = 1

This is of the form \frac{x^2}{a^2}-\frac{y^2}{b^2} = 1   where, a2 = 9, b2 = 12 i.e., a = 3 and b = √12

Eccentricity is given by:

e = \sqrt{1+\frac{b^2}{a^2}}

= \sqrt{1+\frac{12}{9}}

Eccentricity = \frac{7}{3}.

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

The length of latus-rectum is given as= 2b2/a = 2(12)/3 = 24/3 = 8

(iv) 3x2 – y2 = 4

Solution:

Given: 3x2 – y2 = 4

⇒ \frac{3x^2}{4}-\frac{y^2}{4} = 1

⇒ \frac{x^2}{\frac{4}{3}}-\frac{y^2}{4} = 1

⇒ \frac{x^2}{(\frac{2}{\sqrt3})^2}-\frac{y^2}{(2)^2} = 1



This is of the form \frac{x^2}{a^2}-\frac{y^2}{b^2} = 1   where, a = \frac{2}{\sqrt3}   and b = 2

Eccentricity is given by:

e = \sqrt{1+\frac{b^2}{a^2}}

= \sqrt{1+\frac{4}{3}}

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) 2x2 – 3y2 = 5

Solution:

Given: 2x2 – 3y2 = 5

⇒ \frac{2x^2}{5}-\frac{3y^2}{5} = 1

⇒ \frac{x^2}{\frac{5}{2}}-\frac{y^2}{\frac{5}{3}} = 1

⇒ \frac{x^2}{(\sqrt{\frac{5}{2})}^2}-\frac{y^2}{(\sqrt{\frac{5}{3})}^2} = 1

This is of the form \frac{x^2}{a^2}-\frac{y^2}{b^2}=1   where, a = \sqrt{\frac{5}{2}}   and b = \sqrt{\frac{5}{3}}

Eccentricity is given by:

e = \sqrt{1+\frac{b^2}{a^2}}

= \sqrt{1+\frac{2}{3}}

Eccentricity = \sqrt{\frac{5}{3}}.

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

or, (±ae, 0) = (±\frac{5}{\sqrt6}, 0)



The length of latus-rectum is given as: 2b2/a\frac{10\sqrt2}{3\sqrt5}

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

Solution:

Given: 25x2 – 36y2 = 225

⇒ \frac{25x^2}{225}-\frac{36y^2}{225} = 1

⇒ \frac{x^2}{9}-\frac{4y^2}{25} = 1

⇒ \frac{x^2}{3^2}-\frac{y^2}{(\frac{5}{2})^2} = 1

This is of the form \frac{x^2}{a^2}-\frac{y^2}{b^2}=1   where, a = 3 and b = 5/2

Eccentricity is given by:

e = \sqrt{1+\frac{b^2}{a^2}}

= \sqrt{1+\frac{25}{36}}

= \frac{\sqrt61}{6}

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

(±ae, 0) = (± √61/2, 0)

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)

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

(i) 16x2 – 9y2 + 32x + 36y – 164 = 0

Solution:

Given: 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 \frac{x^2}{a^2}-\frac{y^2}{b^2}=1   where, a = 3 and b = 4.

Eccentricity is given by:

e = \sqrt{1+\frac{b^2}{a^2}}

= \sqrt{1+\frac{16}{9}}

= \frac{5}{3}

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) x2 – y2 + 4x = 0

Solution:

Given:  x2 – y2 + 4x = 0.

⇒ x2 – y2 + 4x = 0

⇒ x2 + 4x + 4 – y2 – 4 = 0

⇒ (x + 2)2 – y2 = 4

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

The obtained equation is of the form \frac{x^2}{a^2}-\frac{y^2}{b^2}=1   where, a = 2 and b = 2

Eccentricity is given by:

e = \sqrt{1+\frac{b^2}{a^2}}

= \sqrt{1+\frac{4}{4}}

= \sqrt{2}

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) x2 – 3y2 – 2x = 8

Solution:

Given: x2 – 3y2 – 2x = 8.

⇒ x2 – 3y2 – 2x = 8

⇒ x2 – 2x + 1 – 3y2 – 1 = 8

⇒ (x – 1)2 – 3y2 = 9

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

The obtained equation is of the form \frac{x^2}{a^2}-\frac{y^2}{b^2}=1   where, a = 3 and b = √3

Eccentricity is given by:



e = \sqrt{1+\frac{b^2}{a^2}}

= \sqrt{1+\frac{3}{9}}

= \frac{2\sqrt{3}}{3}

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 \frac{x^2}{a^2}-\frac{y^2}{b^2}=1           …..(1)

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 \frac{x^2}{32}-\frac{y^2}{32}=1  

⇒ x2 – y2 = 32

∴ The Equation of hyperbola is x2 – y2 = 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 \frac{x^2}{a^2}-\frac{y^2}{b^2}=1              …..(1)

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

⇒ a2 = 144/4 = 36

The Equation of hyperbola is given as \frac{x^2}{36}-\frac{y^2}{\frac{25}{4}}=1

⇒ \frac{x^2}{36}-\frac{4y^2}{25}=1

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

⇒ b2 = 49/4



The Equation of hyperbola is given as \frac{x^2}{a^2}-\frac{y^2}{b^2}=1

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

\frac{3^2}{a^2}-\frac{(-2)^2}{b^2}=1

⇒ \frac{9}{a^2}-\frac{4}{\frac{49}{4}}=1

⇒ a2 = 441/65

The equation of hyperbola is given as: \frac{x^2}{\frac{441}{65}}-\frac{y^2}{\frac{49}{4}}=1

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

\frac{(x-1)^2}{a^2}-\frac{(y-4)^2}{b^2} = 1

Distance between the foci = 2ae

⇒ 2ae = 10

⇒ a = 5/2

⇒ a2 = 25/4

Since, b2 = a2(e2 – 1)

⇒ b2 = 75/4

Putting the values in the equation, we get

\frac{(x-1)^2}{\frac{25}{4}}-\frac{(y-4)^2}{\frac{75}{4}} = 1

⇒ 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 : 

\frac{(x-4)^2}{a^2}-\frac{(y+1)^2}{b^2} = 1

Distance between vertices = 2ae

⇒ 24 = 2a

⇒ a = 12

⇒ a2 = 144 and e2 = 169/144

Since, b2 = a2(e2 – 1)

⇒ b2 = 25

Putting the values in the equation, we get

\frac{(x-1)^2}{144}-\frac{(y-4)^2}{25} = 1

⇒ 25x2 – 144y2 – 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 :

\frac{(x-6)^2}{a^2}-\frac{(y-2)^2}{b^2} = 1

Distance between the foci = 2ae

⇒ 2ae = 4

⇒ a = 1



Since, b2 = a2(e2 – 1)

⇒ b2 =3

Putting the values in the equation, we get

\frac{(x-6)^2}{1}-\frac{(y-2)^2}{3} = 1

⇒ 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

⇒ b2 = 49

and, be = 28/3

⇒ e = 4/3 ⇒ e2 = 16/9

Now, a2 = b2(e2-1)

⇒ a2 = 343/9

The equation becomes:

\frac{x^2}{\frac{343}{9}}-\frac{y^2}{49} = 1

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

⇒ b2/a2 = 9/16

Now, e = \sqrt{1+\frac{b^2}{a^2}}

= \sqrt{\frac{25}{16}}

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 :

\frac{(x-3)^2}{a^2}-\frac{(y-2)^2}{b^2} = 1

Distance between 2 vertices = 2a

 ⇒ a = 1

and, e = 2

 b2 = a2(e2 – 1)

⇒ b2 = 3

The equation becomes:

\frac{(x-3)^2}{1}-\frac{(y-2)^2}{3} = 1

⇒ 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 = CP2.

Solution:

Given: a = b

Equation becomes: x2 – y2 = a2

e = \sqrt{2} , C = (0,0), S = (\sqrt{2}a,0)  and S' = (-\sqrt{2}a,0)

SP. S’P = 4a4 + 4a2(a2 + b2) + (a2 + b2)2 – 8a2b2

= (a2 + b2)2 = CP

Hence, SP.S’P = CP2.

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 :

\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1

Distance between the foci = 2ae

⇒ a = 2

⇒ a2 = 4

e = 3/2

Since, b2 = a2(e2 – 1)

⇒ b2 = 5



Putting the values in the equation, we get

\frac{x^2}{4}-\frac{y^2}{5} = 1.

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

Solution:

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

⇒ b = 4

⇒ b2 = 16

and, be = 2/3

⇒ e = 2/3 ⇒ e2 = 4/9

Now, a2 = b2(e2-1)

⇒ a2 = 343/9

The equation becomes:

\frac{x^2}{\frac{343}{9}}-\frac{y^2}{49} = 1

Question 12. Find the equation when the distance between foci is 16 and eccentricity is \sqrt{2}.

Solution:

Distance between foci = 2ae = 16

⇒ 2a(\sqrt{2}) = 16

⇒ a = \frac{16}{2\sqrt{2}}

⇒ a = 4\sqrt{2}

e = \frac{\sqrt{a^2+b^2}}{a}

or, b2 = 32

Equation becomes: x2 – y2 = 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) = \sqrt{(x-4)^2+y^2}

Distance of P from (-4,0) = \sqrt{(x+4)^2+y^2}

Given: 

\sqrt{(x-4)^2+y^2} - \sqrt{(x+4)^2+y^2} = 2

⇒ \sqrt{(x-4)^2+y^2} = \sqrt{(x+4)^2+y^2} + 2

Squaring both sides, we have

-4x-1 = \sqrt{(x+4)^2+y^2}

⇒ 15x2 – y2 = 15.

Thus, P represents a hyperbola.

Attention reader! Don’t stop learning now. Join the First-Step-to-DSA Course for Class 9 to 12 students , specifically designed to introduce data structures and algorithms to the class 9 to 12 students




My Personal Notes arrow_drop_up
Recommended Articles
Page :