Question

Find the equation of the hyperbola referred to its principal axes:

whose foci are at (±2, 0) and eccentricity `3/2`

Answer 1

Let the required equation of hyperbola be

`x^2/"a"^2 - y^2/"b"^2` = 1.    ...(i)

Given, eccentricity(e) = `3/2`

Co-ordinates of foci are (±ae, 0).

Given co-ordinates of foci are (±2, 0).

∴ ae = 2

∴ `"a"(3/2)` = 2

∴ a = `4/3`

∴ a² = `16/9`

Now, b² = a²(e² – 1)

∴ b² = a²e² – a²

= `4 - 16/9`

= `20/9`

∴ The required equation of hyperbola is

`x^2/((16/9)) - y^2/((20/9))` = 1, i.e., `(9x^2)/16 - (9y^2)/20` = 1.