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

Chapter 27 of RD Sharma's Class 11 mathematics textbook focuses on the Hyperbola, a fundamental conic section in analytical geometry. Exercise 27.1 specifically introduces students to the basic concepts, definitions, and standard forms of hyperbolas.

The problems in this exercise are designed to build proficiency in manipulating hyperbolic equations, recognizing the effects of transformations on hyperbolas, and applying these concepts to solve real-world problems.

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

⇒ 12x2 - 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

⇒ 3x2 - 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

(v) vertices are at (±6, 0) and one of the directrices is x = 4.

Solution:

It is given that the vertices of the hyperbola are (±6, 0).

=> a = 6

=> a² = 36

Now, x = 4

=> a/e = 4

=> 6/e = 4

=> e = 3/2

Now we know,

(ae)² = a² + b²

(6 × (3/2))² = 6² + b²

b² = 81 - 36

b² = 45

The equation becomes,

\frac{x^2}{36} - \frac{y^2}{45} = 1

(vi) Whose foci are at (± 2, 0) and eccentricity is 3/2. 

Solution:

We have the foci given as, (± 2, 0).

Here e = 3/2. We know,

ae = 2

=> a = 2/e

=> a = 2/(3/2)

=> a = 4/3

Now we know,

(ae)² = a² + b²

(2)² = (4/3)² + b²

b² = 4 - 16/3

b² = 20/9

So the equation becomes,,

=> \frac{9 x^2}{16} - \frac{9 y^2}{20} = 1

=> \frac{x^2}{4} - \frac{y^2}{5} = \frac{4}{9}

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.

Summary

Exercise 27.1 in RD Sharma's Class 11 textbook provides a comprehensive introduction to hyperbolas, covering their standard forms, key characteristics, and geometric properties. Through a variety of problem types, students learn to identify, analyze, and construct hyperbolas in different contexts. The exercise emphasizes the importance of understanding the relationships between algebraic expressions and geometric representations of hyperbolas. By working through these problems, students develop crucial skills in algebraic manipulation, geometric visualization, and analytical reasoning. This foundational knowledge of hyperbolas is essential for advancing to more complex topics in conic sections and analytical geometry. The practical applications of hyperbolas in fields such as physics, engineering, and astronomy are also highlighted, demonstrating the relevance of these mathematical concepts beyond the classroom. Mastery of the concepts presented in this exercise prepares students for more advanced studies in mathematics and its applications in various scientific and technical fields.