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

• Last Updated : 25 Jun, 2022

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

The equation of the directrix of a hyperbola â‡’ x â€“ y + 3 = 0.

Let â€˜Mâ€™ be the point on directrix and P(x, y) be any point of the hyperbola.

We know, e = PF/PM â‡’ PF2 = e2PM2

â‡’ 2{x2 + 1 + 2x + y2 + 1 â€“ 2y} = 9{x2 + y2+ 9 + 6x â€“ 6y â€“ 2xy}

â‡’ 2x2 + 2 + 4x + 2y2 + 2 â€“ 4y = 9x2 + 9y2+ 81 + 54x â€“ 54y â€“ 18xy

â‡’ 2x2 + 4 + 4x + 2y2â€“ 4y â€“ 9x2 â€“ 9y2 â€“ 81 â€“ 54x + 54y + 18xy = 0

â‡’ â€“ 7x2 â€“ 7y2 â€“ 50x + 50y + 18xy â€“ 77 = 0

â‡’ 7(x2 + y2) â€“ 18xy + 50x â€“ 50y + 77 = 0

âˆ´The equation of hyperbola is 7(x2 + y2) â€“ 18xy + 50x â€“ 50y + 77 = 0.

### (i) focus is (0, 3), directrix is x + y â€“ 1 = 0 and eccentricity = 2

Solution:

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

Let â€˜Mâ€™ be the point on directrix and P(x, y) be any point of the hyperbola.

We know, e = PF/PM â‡’ PF2 = e2PM2

â‡’ 2{x2 + y2 + 9 â€“ 6y} = 4{x2 + y2 + 1 â€“ 2x â€“ 2y + 2xy}

â‡’ 2x2 + 2y2 + 18 â€“ 12y â€“ 4x2 â€“ 4y2 â€“ 4 â€“ 8x + 8y â€“ 8xy = 0

â‡’ â€“ 2x2 â€“ 2y2 â€“ 8x â€“ 4y â€“ 8xy + 14 = 0

â‡’ â€“2(x2 + y2 â€“ 4x + 2y + 4xy â€“ 7) = 0

â‡’ x2 + y2 â€“ 4x + 2y + 4xy â€“ 7 = 0

âˆ´The equation of hyperbola is x2 + y2 â€“ 4x + 2y + 4xy â€“ 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 â€˜Mâ€™ be the point on directrix and P(x, y) be any point of the hyperbola.

We know, e = PF/PM â‡’ PF2 = e2PM2

â‡’ 25{x2 + 1 â€“ 2x + y2 + 1 â€“ 2y} = 4{9x2 + 16y2+ 64 + 24xy + 64y + 48x}

â‡’ 25x2 + 25 â€“ 50x + 25y2 + 25 â€“ 50y = 36x2 + 64y2 + 256 + 96xy + 256y + 192x

â‡’ 25x2 + 25 â€“ 50x + 25y2 + 25 â€“ 50y â€“ 36x2 â€“ 64y2 â€“ 256 â€“ 96xy â€“ 256y â€“ 192x = 0

â‡’ â€“ 11x2 â€“ 39y2 â€“ 242x â€“ 306y â€“ 96xy â€“ 206 = 0

â‡’ 11x2 + 96xy + 39y2 + 242x + 306y + 206 = 0

âˆ´The equation of hyperbola is 11x2 + 96xy + 39y2 + 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 â€˜Mâ€™ be the point on directrix and P(x, y) be any point of the hyperbola.

We know, e = PF/PM â‡’ PF2 = e2PM2

â‡’ 5{x2 + 1 â€“ 2x + y2 + 1 â€“ 2y} = 3{4x2 + y2+ 1 + 4xy â€“ 2y â€“ 4x}

â‡’ 5x2 + 5 â€“ 10x + 5y2 + 5 â€“ 10y = 12x2 + 3y2 + 3 + 12xy â€“ 6y â€“ 12x

â‡’ 5x2 + 5 â€“ 10x + 5y2 + 5 â€“ 10y â€“ 12x2 â€“ 3y2 â€“ 3 â€“ 12xy + 6y + 12x = 0

â‡’ â€“ 7x2 + 2y2 + 2x â€“ 4y â€“ 12xy + 7 = 0

â‡’ 7x2 + 12xy â€“ 2y2 â€“ 2x + 4yâ€“ 7 = 0

âˆ´The equation of hyperbola is 7x2 + 12xy â€“ 2y2 â€“ 2x + 4yâ€“ 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 â€˜Mâ€™ be the point on directrix and P(x, y) be any point of the hyperbola.

By using the formula, e = PF/PM â‡’ PF2 = e2PM2

â‡’ 13{x2 + 4 â€“ 4x + y2 + 1 + 2y} = 4{4x2 + 9y2 + 1 + 12xy â€“ 6y â€“ 4x}

â‡’ 13x2 + 52 â€“ 52x + 13y2 + 13 + 26y = 16x2 + 36y2 + 4 + 48xy â€“ 24y â€“ 16x

â‡’ 13x2 + 52 â€“ 52x + 13y2 + 13 + 26y â€“ 16x2 â€“ 36y2 â€“ 4 â€“ 48xy + 24y + 16x = 0

â‡’ â€“ 3x2 â€“ 23y2 â€“ 36x + 50y â€“ 48xy + 61 = 0

â‡’ 3x2 + 23y2 + 48xy + 36x â€“ 50yâ€“ 61 = 0

âˆ´The equation of hyperbola is 3x2 + 23y2 + 48xy + 36x â€“ 50yâ€“ 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 â€˜Mâ€™ be the point on directrix and P(x, y) be any point of the hyperbola.

By using the formula, e = PF/PM â‡’ PF2 = e2PM2

â‡’ 45{x2 + a2 â€“ 2ax + y2} = 16{4x2 + y2 + a2 â€“ 4xy â€“ 2ay + 4ax}

â‡’ 45x2 + 45a2 â€“ 90ax + 45y2 = 64x2 + 16y2 + 16a2 â€“ 64xy â€“ 32ay + 64ax

â‡’ 45x2 + 45a2 â€“ 90ax + 45y2 â€“ 64x2 â€“ 16y2 â€“ 16a2 + 64xy + 32ay â€“ 64ax = 0

â‡’ 19x2 â€“ 29y2 + 154ax â€“ 32ay â€“ 64xy â€“ 29a2 = 0

âˆ´The equation of hyperbola is 19x2 â€“ 29y2 + 154ax â€“ 32ay â€“ 64xy â€“ 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 â€˜Mâ€™ be the point on directrix and P(x, y) be any point of the hyperbola.

By using the formula, e = PF/PM â‡’ PF2 = e2PM2

â‡’ x2 + 4 â€“ 4x + y2 + 4 â€“ 4y = 2{x2 + y2 + 81 + 2xy â€“ 18y â€“ 18x}

â‡’ x2 â€“ 4x + y2 + 8 â€“ 4y = 2x2 + 2y2 + 162 + 4xy â€“ 36y â€“ 36x

â‡’ x2 â€“ 4x + y2 + 8 â€“ 4y â€“ 2x2 â€“ 2y2 â€“ 162 â€“ 4xy + 36y + 36x = 0

â‡’ â€“ x2 â€“ y2 + 32x + 32y + 4xy â€“ 154 = 0

â‡’ x2 + 4xy + y2 â€“ 32x â€“ 32y + 154 = 0

âˆ´The equation of hyperbola is x2 + 4xy + y2 â€“ 32x â€“ 32y + 154 = 0.

### (i) 9x2 â€“ 16y2 = 144

Solution:

Given: 9x2 â€“ 16y2 = 144

This is of the form  where, a2 = 16, b2 = 9 i.e., a = 4 and b = 3

Eccentricity is given by:

Eccentricity

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

Foci = (Â±5, 0)

The equation of directrices is given as:  â‡’ 5x âˆ“ 16 = 0

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

Length of latus rectum= 9/2

### (ii) 16x2 â€“ 9y2 = â€“144

Solution:

Given: 16x2 â€“ 9y2 = â€“144

This is of the form  where, a2 = 9, b2 = 16 i.e., a = 3 and b = 4.

Eccentricity is given by:

Eccentricity =

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: x =  â‡’ 5x âˆ“ 16 = 0.

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

### (iii) 4x2 â€“ 3y2 = 36

Solution:

Given: 4x2 â€“ 3y2 = 36

This is of the form  where, a2 = 9, b2 = 12 i.e., a = 3 and b = âˆš12

Eccentricity is given by:

Eccentricity =

Foci: The coordinates of the foci are (Â±ae, 0)= (Â±ae, 0) = (Â±âˆš21, 0)

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

### (iv) 3x2 â€“ y2 = 4

Solution:

Given: 3x2 â€“ y2 = 4

This is of the form  where,  and b = 2

Eccentricity is given by:

Eccentricity = 2

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 length of latus-rectum is given as:= 2b2/a = 2(4)/[2/âˆš3] = 4âˆš3.

### (v) 2x2 â€“ 3y2 = 5

Solution:

Given: 2x2 â€“ 3y2 = 5

This is of the form  where,  and

Eccentricity is given by:

Eccentricity = .

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

or, (Â±ae, 0) =

The length of latus-rectum is given as: 2b2/a

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

Solution:

Given: 25x2 â€“ 36y2 = 225

This 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± âˆš61/2, 0)

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)

### (i) 16x2 â€“ 9y2 + 32x + 36y â€“ 164 = 0

Solution:

Given: 16x2 â€“ 9y2 + 32x + 36y â€“ 164 = 0.

â‡’ 16x2 + 32x + 16 â€“ 9y2 + 36y â€“ 36 â€“ 16 + 36 â€“ 164 = 0

â‡’ 16(x2 + 2x + 1) â€“ 9(y2 â€“ 4y + 4) â€“ 16 + 36 â€“ 164 = 0

â‡’ 16(x2 + 2x + 1) â€“ 9(y2 â€“ 4y + 4) â€“ 144 = 0

â‡’ 16(x + 1)2 â€“ 9(y â€“ 2)2 = 144

Here, 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±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)

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) x2 â€“ y2 + 4x = 0

Solution:

Given:  x2 â€“ y2 + 4x = 0.

â‡’ x2 â€“ y2 + 4x = 0

â‡’ x2 + 4x + 4 â€“ y2 â€“ 4 = 0

â‡’ (x + 2)2 â€“ y2 = 4

Here, 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 (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)

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

### (iii) x2 â€“ 3y2 â€“ 2x = 8

Solution:

Given: x2 â€“ 3y2 â€“ 2x = 8.

â‡’ x2 â€“ 3y2 â€“ 2x = 8

â‡’ x2 â€“ 2x + 1 â€“ 3y2 â€“ 1 = 8

â‡’ (x â€“ 1)2 â€“ 3y2 = 9

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

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)

âˆ´ The center is (1, 0), eccentricity (e) = 2âˆš3/3, Foci = (1 Â± 2âˆš3, 0), Equation of directrix = X = 1Â±9/2âˆš3.

### (i) the distance between the foci = 16 and eccentricity = âˆš2

Solution:

Given: Distance between the foci = 16 and Eccentricity = âˆš2

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

Distance between the foci is 2ae and b2 = a2(e2 â€“ 1)

So, 2ae = 16

â‡’ ae = 16/2

â‡’ aâˆš2 = 8

â‡’ a = 8/âˆš2

â‡’ a2 = 64/2 = 32

We know that, b2 = a2(e2 â€“ 1)

So, b2 = 32[(âˆš2)2 â€“ 1]

= 32(2 â€“ 1)

= 32

The Equation of hyperbola is given as

â‡’ 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

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 b2 = a2(e2 â€“ 1)

Length of conjugate axis is 2b

So, 2b = 5

â‡’ b = 5/2

â‡’ b2 = 25/4

We know that, 2ae = 13

ae = 13/2

â‡’ a2e2 = 169/4

b2 = a2(e2 â€“ 1)

â‡’ b2 = a2e2 â€“ a2

â‡’ 25/4 = 169/4 â€“ a2

â‡’ a2 = 169/4 â€“ 25/4

â‡’ a2 = 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)

Solution:

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

Conjugate axis is 2b

So, 2b = 7

â‡’ b = 7/2

â‡’ b2 = 49/4

The Equation of hyperbola is given as

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

â‡’ a2 = 441/65

The equation of hyperbola is given as:

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

### (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

â‡’ 2ae = 10

â‡’ a = 5/2

â‡’ a2 = 25/4

Since, b2 = a2(e2 â€“ 1)

â‡’ b2 = 75/4

Putting the values in the equation, we get

â‡’ 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

â‡’ 24 = 2a

â‡’ a = 12

â‡’ a2 = 144 and e2 = 169/144

Since, b2 = a2(e2 â€“ 1)

â‡’ b2 = 25

Putting the values in the equation, we get

â‡’ 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:

Distance between the foci = 2ae

â‡’ 2ae = 4

â‡’ a = 1

Since, b2 = a2(e2 â€“ 1)

â‡’ b2 =3

Putting the values in the equation, we get

â‡’ 3×2 – y2 – 36x + 4y + 101 = 0.

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

Solution:

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

â‡’ b = 7

â‡’ b2 = 49

and, be = 28/3

â‡’ e = 4/3 â‡’ e2 = 16/9

Now, a2 = b2(e2-1)

â‡’ a2 = 343/9

The equation becomes:

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

Solution:

It is given that the vertices of the hyperbola are (Â±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 Ã— (3/2))2 = 62 + b2

b2 = 81 – 36

b2 = 45

The equation becomes,

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

Solution:

We have the foci given as, (Â± 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,,

=>

=>

Solution:

Given: 2b = 6a/4

â‡’ b/a = 3/4

â‡’ b2/a2 = 9/16

Now,

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

â‡’ a = 1

and, e = 2

b2 = a2(e2 â€“ 1)

â‡’ b2 = 3

The equation becomes:

â‡’ 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

, C = (0,0),  and

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

= (a2 + b2)2 = CP

Hence, SP.S’P = CP2.

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

Solution:

Equation of the hyperbola is :

Distance between the foci = 2ae

â‡’ a = 2

â‡’ a2 = 4

e = 3/2

Since, b2 = a2(e2 â€“ 1)

â‡’ b2 = 5

Putting the values in the equation, we get

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

Solution:

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

â‡’ b = 4

â‡’ b2 = 16

and, be = 2/3

â‡’ e = 2/3 â‡’ e2 = 4/9

Now, a2 = b2(e2-1)

â‡’ a2 = 343/9

The equation becomes:

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

Solution:

Distance between foci = 2ae = 16

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) =

Distance of P from (-4,0) =

Given:

Squaring both sides, we have

â‡’ 15x2 – y2 = 15.

Thus, P represents a hyperbola.

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