# Examples on Hyperbola

**For the following hyperbola, find eccentricity, the coordinates of vertices and foci, the lengths of transverse and conjugate axes, and equations of directrices**

**9x**^{2}- 16y^{2}= 144

**Solution**

We can write the given equation as, so that a

^{2}= 16 and b

^{2}= 9.

Since b

^{2}=a

^{2}(e

^{2}-1) we get 9 = 16 (e

^{2}- 1) â‡’ e

^{2}= + 1 = or e = .

The vertices of the hyperbola are (a, 0) = (4, 0) and foci are (ae, 0) = (5, 0). Length of the transverse axis is 2a = 8 and the length of the conjugate axis is 2b = 6.

Equation directrices are x = = .

**For the following hyperbola, find eccentricity, the coordinates of vertices and foci, the lengths of transverse and conjugate axes, and equations of directrices**

**16x**^{2}- 9y^{2}= 144

**Solution**

We can write the given equation as so that a

^{2}= 9 and b

^{2}= 16. Since b

^{2}= a

^{2 }(e

^{2}- 1), we get 16 = 9 (e

^{2}- 1) or e

^{2}= + 1 = or e = . The vertices of the hyperbola are (a, 0) = (3, 0) and foci are (ae, 0) = (5, 0). Length of the transverse axis is 2a = 6 and the length of conjugate axis is 2b = 8. Equation of directrices are x = = .

**For the following hyperbolas, find eccentricity, the coordinates of vertices and foci, the lengths of transverse and conjugate axes, and equations of directrices**

2x^{2}- 3y^{2}- 6 = 0

**Solution**

We can write the given equation as so that a

^{2}= 3, b

^{2}= 2. Since b

^{2}= a

^{2 }(e

^{2}- 1), we get 2 = 3 (e

^{2}- 1) or e

^{2}= + 1 = or e = . The vertices of the hyperbola are (a, 0) = (, 0) and the foci are (ae, 0) = (, 0). Length of the transverse axis is 2a = 2 and the conjugate axes is 2b = 2. Equations of the directrices are x = = .

**For the following hyperbola, find eccentricity, the coordinates of vertices and foci, the lengths of transverse and conjugate axes, and equations of directrices**

**Solution**

The given equation can be written as so that a

^{2}= 4, b

^{2}= 1. Since b

^{2}= a

^{2 }(e

^{2}- 1) we get 1 = 4(e

^{2}- 1) or e

^{2}= + 1 = or e = . The vertices of the hyperbola are ( a, 0) = ( 2, 0) and the foci are ( ae, 0) = ( , 0). Length of the transverse axis is 2a = 4 and that of conjugate axis is 2b = 2. Equation of the directrices are x = = .

- Find the equation of the hyperbola with
- Vertices (5, 0), foci (7, 0)
- Vertices (0, 7), e =

**Solution**

- Since vertices lie on x-axis, the transverse axis coincides with x-axis. Since the mid-point of vertices is the origin, conjugate axis coincides with the y-axis. Therefore, equation of the hyperbola must be of the form.As vertices are (5, 0), a = 5. Also, ae = 7, thus, b
^{2}= a^{2}(e^{2}- 1) = 7^{2}- 5^{2}= 24.Hence, equation of the required hyperbola is

- Since the vertices lie on the y-axis, the transverse axis coincides with y-axis. Since mid-point of the vertices is the origin, the conjugate axis coincides with x-axis. Therefore, equation of the hyperbola must be of the formSince vertices are (0, 7), a = 7.

^{2}= a

^{2}(e

^{2}- 1) = 49 ( - 1) = .

Hence, equation of the required hyperbola is or 7y

^{2}-9x

^{2}= 343.

- Find the equation of the hyperbola with
- Vertices (0, 6), e =
- Foci (0, ), passing through (2, 3)

**Solution**

- The equation of hyperbola must be of the form .

We have a = 6, e = , therefore b^{2}= a^{2}(e^{2}- 1) = 36 ( - 1) = 64.

Hence, equation of the required hyperbola is or 16y

^{2}- 9x^{2}= 576.- In this case also, equation of the hyperbola must be of the form (1)

Since ae = , b^{2}= a^{2}(e^{2}- 1) = 10 - a^{2}. Therefore we can write (1) (2)

Since (2) passes through (2,3), we get

^{2}) - 4a

^{2}= a

^{2}(10 - a

^{2})

or 90 - 13a

^{2}= 10a

^{2}- a

^{4}

or a

^{4}- 23a

^{2}+ 90 = 0

â‡’ (a

^{2}- 18) (a

^{2}- 5) = 0 â‡’ a

^{2}= 18 or a

^{2}= 5

If a

^{2}= 18, then b

^{2}= 10 - a

^{2}< 0.

Therefore, a

^{2}â‰ 18. Thus a

^{2}= 5. Putting this value in (2),

we get or x

^{2}- y

^{2}+ 5 = 0.

- Find the equation of the hyperbola with
- Directrices x = 4, foci ( 6, 0)
- Distance between foci 14, e =

**Solution**

- As the foci are (6, 0), the transverse axis is the x-axis and centre is at the origin. We have = 4, ae = 6.Therefore, a
^{2}= 24.

Also, b^{2}= a^{2}(e^{2}- 1) = a^{2}e^{2}- a^{2}= 36 - 24 = 12.

Hence the equation of required hyperbola is

- We have 2ae = 14, e = , therefore, a = .

Also b^{2}= a^{2}(e^{2}- 1) = a^{2}e^{2}- a^{2}= 49 - = .

- Find the equation of the set of all points such that the difference of their distances from F
_{1}(4, 0) and F_{2}(-4, 0) is always equal to 2.

**Solution**(i) If P (x, y) is any point of the hyperbola, then

|PF

_{1}- PF

_{2}| = 2

â‡’ PF

_{1}- PF

_{2}= 2 â‡’ PF

_{1}= PF

_{2}2

Squaring both the sides, we get 16x

^{2}+ 8x + 1 = x

^{2}+ 8x + 16 + y

^{2}

â‡’ 15x

^{2}- y

^{2}= 15

This is an equation of required hyperbola.

- Find the equation of the hyperbola whose directrix is 2x + y = 1, focus is (1, 2) and eccentricity is .

**Solution**

If P(x, y) is any point on the hyperbola, then

Its distance from (1, 2) = e {its distance from 2x + y = 1}

By crossmultiplying and squaring we get,

â‡’ 5 [x

^{2}- 2x + 1 + y

^{2}- 4y + 4] = 3 [4x

^{2}+ y

^{2}+ 4xy - 4x - 2y + 1]

â‡’ 7x

^{2}- 2y

^{2}+ 12xy - 2x + 14y - 22 = 0.

- Find the equation of the hyperbola whose eccentricity is focus is (3, 0) and directrix is 4x - 3y = 3.

**Solution**

If P(x, y) is any point on the hyperbola, then

its distance from (3, 0) = e {its distance from 4x - 3y = 3}

â‡’ 16 [x

^{2}- 6x + 9 + y

^{2}] = 16x

^{2}- 24xy +9y

^{2}- 24x + 18y + 9