Question

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

Answer 1

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