1.

Find the equation of hyperbola whose foci are (±5, 0) and the conjugate axis is of the length 8. Also, find its eccentricity.

Answer»

Foci of the equation is in the form (±c, 0)

Equation of the hyperbola will be of the form: x2/a2 – y2/b2 = 1

Whose foci at (±c, 0) and conjugate axis is 2b

On comparing, we get

2b = 8 ⇒ b = 4

And c = 5

Again, We know, c2 = a2 + b2

25 = 16 + a2

or a2 = 9

Equation (1)⇒

x2/9 – y2/16 = 1

Which is required equation.

Again,

Eccentricity (e) = c/a = 5/3



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