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

**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‡’ PF

^{2}= e^{2}PM^{2}â‡’ 2{x

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

^{2}+ 2 + 4x + 2y^{2}+ 2 â€“ 4y = 9x^{2}+ 9y^{2}+ 81 + 54x â€“ 54y â€“ 18xyâ‡’ 2x

^{2}+ 4 + 4x + 2y^{2}â€“ 4y â€“ 9x^{2}â€“ 9y^{2}â€“ 81 â€“ 54x + 54y + 18xy = 0â‡’ â€“ 7x

^{2}â€“ 7y^{2}â€“ 50x + 50y + 18xy â€“ 77 = 0â‡’ 7(x

^{2}+ y^{2}) â€“ 18xy + 50x â€“ 50y + 77 = 0

âˆ´The equation of hyperbola is 7(x^{2}+ y^{2}) â€“ 18xy + 50x â€“ 50y + 77 = 0.

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

**(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‡’ PF

^{2}= e^{2}PM^{2}â‡’ 2{x

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

^{2}+ 2y^{2}+ 18 â€“ 12y â€“ 4x^{2}â€“ 4y^{2}â€“ 4 â€“ 8x + 8y â€“ 8xy = 0â‡’ â€“ 2x

^{2 }â€“ 2y^{2}â€“ 8x â€“ 4y â€“ 8xy + 14 = 0â‡’ â€“2(x

^{2}+ y^{2}â€“ 4x + 2y + 4xy â€“ 7) = 0â‡’ x

^{2}+ y^{2}â€“ 4x + 2y + 4xy â€“ 7 = 0

âˆ´The equation of hyperbola is x^{2}+ y^{2}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‡’ PF

^{2}= e^{2}PM^{2}â‡’ 25{x

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

^{2}+ 25 â€“ 50x + 25y^{2}+ 25 â€“ 50y = 36x^{2}+ 64y^{2}+ 256 + 96xy + 256y + 192xâ‡’ 25x

^{2}+ 25 â€“ 50x + 25y^{2}+ 25 â€“ 50y â€“ 36x^{2}â€“ 64y^{2}â€“ 256 â€“ 96xy â€“ 256y â€“ 192x = 0â‡’ â€“ 11x

^{2}â€“ 39y^{2}â€“ 242x â€“ 306y â€“ 96xy â€“ 206 = 0â‡’ 11x

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

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

We know, e = PF/PM â‡’ PF

^{2}= e^{2}PM^{2}â‡’ 5{x

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

^{2}+ 5 â€“ 10x + 5y^{2}+ 5 â€“ 10y = 12x^{2}+ 3y^{2}+ 3 + 12xy â€“ 6y â€“ 12xâ‡’ 5x

^{2}+ 5 â€“ 10x + 5y^{2}+ 5 â€“ 10y â€“ 12x^{2}â€“ 3y^{2}â€“ 3 â€“ 12xy + 6y + 12x = 0â‡’ â€“ 7x

^{2}+ 2y^{2}+ 2x â€“ 4y â€“ 12xy + 7 = 0â‡’ 7x

^{2}+ 12xy â€“ 2y^{2}â€“ 2x + 4yâ€“ 7 = 0

âˆ´The equation of hyperbola is 7x^{2}+ 12xy â€“ 2y^{2 }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‡’ PF

^{2}= e^{2}PM^{2}â‡’ 13{x

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

^{2}+ 52 â€“ 52x + 13y^{2}+ 13 + 26y = 16x^{2}+ 36y^{2}+ 4 + 48xy â€“ 24y â€“ 16xâ‡’ 13x

^{2}+ 52 â€“ 52x + 13y^{2}+ 13 + 26y â€“ 16x^{2}â€“ 36y^{2 }â€“ 4 â€“ 48xy + 24y + 16x = 0â‡’ â€“ 3x

^{2}â€“ 23y^{2}â€“ 36x + 50y â€“ 48xy + 61 = 0â‡’ 3x

^{2}+ 23y^{2}+ 48xy + 36x â€“ 50yâ€“ 61 = 0

âˆ´The equation of hyperbola is 3x^{2}+ 23y^{2}+ 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‡’ PF

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

â‡’ 45{x

^{2}+ a^{2 }â€“ 2ax + y2} = 16{4x^{2}+ y^{2 }+ a^{2}â€“ 4xy â€“ 2ay + 4ax}â‡’ 45x

^{2}+ 45a^{2}â€“ 90ax + 45y^{2}= 64x^{2}+ 16y^{2}+ 16a^{2}â€“ 64xy â€“ 32ay + 64axâ‡’ 45x

^{2}+ 45a^{2}â€“ 90ax + 45y^{2}â€“ 64x^{2}â€“ 16y^{2}â€“ 16a^{2}+ 64xy + 32ay â€“ 64ax = 0â‡’ 19x

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

âˆ´The equation of hyperbola is 19x^{2}â€“ 29y^{2}+ 154ax â€“ 32ay â€“ 64xy â€“ 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 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‡’ PF

^{2}= e^{2}PM^{2}â‡’ x

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

^{2}â€“ 4x + y^{2}+ 8 â€“ 4y = 2x^{2}+ 2y^{2}+ 162 + 4xy â€“ 36y â€“ 36xâ‡’ x

^{2}â€“ 4x + y^{2}+ 8 â€“ 4y â€“ 2x^{2 }â€“ 2y^{2}â€“ 162 â€“ 4xy + 36y + 36x = 0â‡’ â€“ x

^{2}â€“ y^{2}+ 32x + 32y + 4xy â€“ 154 = 0â‡’ x

^{2}+ 4xy + y^{2}â€“ 32x â€“ 32y + 154 = 0

âˆ´The equation of hyperbola is x^{2}+ 4xy + y^{2}â€“ 32x â€“ 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} â€“ 16y^{2} = 144

^{2}â€“ 16y

^{2}= 144

**Solution:**

Given: 9x

^{2}â€“ 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 (Â±ae, 0)

Foci = (Â±5, 0)

The equation of directrices is given as:â‡’5x âˆ“ 16 = 0The length of latus-rectum is given as: 2b

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

Length of latus rectum= 9/2

**(ii) 16x**^{2} â€“ 9y^{2} = â€“144

^{2}â€“ 9y

^{2}= â€“144

**Solution:**

Given: 16x

^{2}â€“ 9y^{2}= â€“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, 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: 2a^{2}/b = 2(9)/4 = 9/2.

**(iii) 4x**^{2} â€“ 3y^{2} = 36

^{2}â€“ 3y

^{2}= 36

**Solution:**

Given: 4x

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

^{2}= 9, b^{2}= 12 i.e., a = 3 and b = âˆš12Eccentricity 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= 2b^{2}/a = 2(12)/3 = 24/3 = 8

**(iv) 3x**^{2} â€“ y^{2} = 4

^{2}â€“ y

^{2}= 4

**Solution:**

Given: 3x

^{2}â€“ y^{2}= 4This 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) 2x**^{2} â€“ 3y^{2} = 5

^{2}â€“ 3y

^{2}= 5

**Solution:**

Given: 2x

^{2}â€“ 3y^{2}= 5This 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: 2b=^{2}/a

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

^{2}â€“ 36y

^{2}= 225.

**Solution:**

Given: 25x

^{2}â€“ 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 (Â±ae, 0)

(Â±ae, 0) = (Â± âˆš61/2, 0)

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

^{2}/a

âˆ´ Transverse axis = 6, conjugate axis = 5, e = âˆš61/6, LR = 25/6, foci = (Â± âˆš61/2, 0)

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

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

^{2}â€“ 9y

^{2}+ 32x + 36y â€“ 164 = 0

**Solution:**

Given: 16x

^{2}â€“ 9y^{2}+ 32x + 36y â€“ 164 = 0.â‡’ 16x

^{2}+ 32x + 16 â€“ 9y^{2}+ 36y â€“ 36 â€“ 16 + 36 â€“ 164 = 0â‡’ 16(x

^{2}+ 2x + 1) â€“ 9(y^{2}â€“ 4y + 4) â€“ 16 + 36 â€“ 164 = 0â‡’ 16(x

^{2}+ 2x + 1) â€“ 9(y^{2}â€“ 4y + 4) â€“ 144 = 0â‡’ 16(x + 1)

^{2}â€“ 9(y â€“ 2)^{2}= 144Here, 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) x**^{2} â€“ y^{2} + 4x = 0

^{2}â€“ y

^{2}+ 4x = 0

**Solution:**

Given: x

^{2}â€“ y^{2}+ 4x = 0.â‡’ x

^{2}â€“ y^{2 }+ 4x = 0â‡’ x

^{2}+ 4x + 4 â€“ y^{2}â€“ 4 = 0â‡’ (x + 2)

^{2}â€“ 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 (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) x**^{2} â€“ 3y^{2} â€“ 2x = 8

^{2}â€“ 3y

^{2}â€“ 2x = 8

**Solution:**

Given: x

^{2}â€“ 3y^{2}â€“ 2x = 8.â‡’ x

^{2}â€“ 3y^{2}â€“ 2x = 8â‡’ x

^{2}â€“ 2x + 1 â€“ 3y^{2}â€“ 1 = 8â‡’ (x â€“ 1)

^{2}â€“ 3y^{2}= 9Here, 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.

**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 = 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 b

^{2}= a^{2}(e^{2}â€“ 1)So, 2ae = 16

â‡’ ae = 16/2

â‡’ aâˆš2 = 8

â‡’ a = 8/âˆš2

â‡’ a

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

^{2}= a^{2}(e^{2}â€“ 1)So, b

^{2}= 32[(âˆš2)2 â€“ 1]= 32(2 â€“ 1)

= 32

The Equation of hyperbola is given as

â‡’ x

^{2}â€“ y^{2}= 32

âˆ´ The Equation of hyperbola is x^{2}â€“ 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}â€“ 1)Length of conjugate axis is 2b

So, 2b = 5

â‡’ b = 5/2

â‡’ b

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

ae = 13/2

â‡’ a

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

^{2}= a^{2}(e^{2 }â€“ 1)â‡’ b

^{2}= a^{2}e^{2}â€“ a^{2}â‡’ 25/4 = 169/4 â€“ a

^{2}â‡’ a

^{2}= 169/4 â€“ 25/4â‡’ a

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

âˆ´ The Equation of hyperbola is 25x^{2}â€“ 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

â‡’ b = 7/2

â‡’ b

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

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

â‡’ a

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

âˆ´ The Equation of hyperbola is 65x^{2}â€“ 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

â‡’ 2ae = 10

â‡’ a = 5/2

â‡’ a

^{2}= 25/4Since, b

^{2}= a^{2}(e^{2}â€“ 1)â‡’ b

^{2}= 75/4Putting 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‡’ a

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

^{2}= a^{2}(e^{2}â€“ 1)â‡’ b

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

â‡’ 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

â‡’ 2ae = 4

â‡’ a = 1

Since, b

^{2}= a^{2}(e^{2}â€“ 1)â‡’ b

^{2}=3Putting 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‡’ b

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

â‡’ e = 4/3 â‡’ e

^{2}= 16/9Now, a

^{2}= b^{2}(e^{2}-1)â‡’ a

^{2}= 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

=> a

^{2}= 36Now, x = 4

=> a/e = 4

=> 6/e = 4

=> e = 3/2

Now we know,

(ae)

^{2}= a^{2}+ b^{2}(6 Ã— (3/2))

^{2}= 6^{2}+ b^{2}b

^{2}= 81 – 36b

^{2}= 45The 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}= a^{2}+ b^{2}(2)

^{2}= (4/3)^{2}+ b^{2}b

^{2}= 4 – 16/3b

^{2}= 20/9So 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

â‡’ b/a = 3/4

â‡’ 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

â‡’ a = 1

and, e = 2

b

^{2}= a^{2}(e^{2}â€“ 1)â‡’ b

^{2}= 3The 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 = 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 (Â±2,0) and foci are (Â±3,0).**

**Solution:**

Equation of the hyperbola is :

Distance between the foci = 2ae

â‡’ a = 2

â‡’ a

^{2}= 4e = 3/2

Since, b

^{2}= a^{2}(e^{2}â€“ 1)â‡’ b

^{2}= 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‡’ b

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

â‡’ e = 2/3 â‡’ e

^{2}= 4/9Now, a

^{2}= b^{2}(e^{2}-1)â‡’ 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

â‡’ 15x

^{2}– y^{2}= 15.

Thus, P represents a hyperbola.

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