# Examples on Hyperbola

1. For the following hyperbola, find eccentricity, the coordinates of vertices and foci, the lengths of transverse and conjugate axes, and equations of directrices
9x2 - 16y2 = 144
Solution
We can write the given equation as, so that a2 = 16 and b2 = 9.
Since b2 =a2 (e2 -1) we get 9 = 16 (e2 - 1) â‡’ e2 = + 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 = = .

1. For the following hyperbola, find eccentricity, the coordinates of vertices and foci, the lengths of transverse and conjugate axes, and equations of directrices
16x2 - 9y2 = 144
Solution
We can write the given equation as so that a2 = 9 and b2 = 16. Since b2 = a2 (e2 - 1), we get 16 = 9 (e2 - 1) or e2 = + 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 = = .

1. For the following hyperbolas, find eccentricity, the coordinates of vertices and foci, the lengths of transverse and conjugate axes, and equations of directrices
2x2 - 3y2 - 6 = 0
Solution
We can write the given equation as so that a2 = 3, b2 = 2. Since b2 = a2 (e2 - 1), we get 2 = 3 (e2 - 1) or e2 = + 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 = = .

1. 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 a2 = 4, b2 = 1. Since b2 = a2 (e2 - 1) we get 1 = 4(e2 - 1) or e2 = + 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 = = .

1. Find the equation of the hyperbola with
1. Vertices (5, 0), foci (7, 0)
2. Vertices (0, 7), e =
Solution
1. 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, b2 = a2 (e2 - 1) = 72 - 52 = 24.Hence, equation of the required hyperbola is

1. 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.

Also, as e = , we get b2 = a2 (e2 - 1) = 49 ( - 1) = .
Hence, equation of the required hyperbola is or 7y2 -9x2 = 343.
1. Find the equation of the hyperbola with
1. Vertices (0, 6), e =
2. Foci (0, ), passing through (2, 3)
Solution
1. The equation of hyperbola must be of the form .
We have a = 6, e = , therefore b2 = a2 (e2 - 1) = 36 ( - 1) = 64.

Hence, equation of the required hyperbola is or 16y2 - 9x2 = 576.
1. In this case also, equation of the hyperbola must be of the form (1)
Since ae = , b2 = a2 (e2 - 1) = 10 - a2. Therefore we can write (1)                               (2)

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

or 9 (10 - a2) - 4a2 = a2 (10 - a2)
or 90 - 13a2 = 10a2 - a4
or a4 - 23a2 + 90 = 0
â‡’ (a2 - 18) (a2 - 5) = 0 â‡’ a2 = 18 or a2 = 5
If a2 = 18, then b2 = 10 - a2 < 0.
Therefore, a2â‰  18. Thus a2 = 5. Putting this value in (2),
we get or x2 - y2 + 5 = 0.
1. Find the equation of the hyperbola with
1. Directrices x = 4, foci ( 6, 0)
2. Distance between foci 14, e =
Solution
1. 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, a2 = 24.
Also, b2 = a2 (e2 - 1) = a2 e2 - a2 = 36 - 24 = 12.

Hence the equation of required hyperbola is
1. We have 2ae = 14, e = , therefore, a = .
Also b2 = a2 (e2 - 1) = a2 e2 - a2 = 49 - = .

Thus, equation of the required hyperbola is
1. Find the equation of the set of all points such that the difference of their distances from F1(4, 0) and F2(-4, 0) is always equal to 2.
Solution

(i) If P (x, y) is any point of the hyperbola, then
|PF1 - PF2| = 2
â‡’ PF1 - PF2 = 2 â‡’ PF1 = PF2 2

Squaring both the sides, we get 16x2 + 8x + 1 = x2 + 8x + 16 + y2
â‡’ 15x2 - y2 = 15
This is an equation of required hyperbola.
1. 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 [x2 - 2x + 1 + y2 - 4y + 4] = 3 [4x2 + y2 + 4xy - 4x - 2y + 1]
â‡’ 7x2 - 2y2 + 12xy - 2x + 14y - 22 = 0.
1. 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 [x2 - 6x + 9 + y2] = 16x2 - 24xy +9y2 - 24x + 18y + 9