Question

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

whose distance between directrices is `8/3` and eccentricity is `3/2`

Answer 1

Let the required equation of hyperbola be `x^2/"a"^2 - y^2/"b"^2` = 1

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

Distance between directrices = `(2"a")/"e"`

Given, distance between directrices = `8/3`

∴ `(2"a")/"e" = 8/3`

∴ `(2"a")/(3/2) = 8/3`

∴ `(4"a")/3 = 8/3`

∴ a = 2

∴ a² = 4

∴ Now, b² = a²(e² – 1)

∴ b² = `4[(3/2)^2 - 1]`

= `4(9/4 - 1)`

= `4(5/4)`

∴ b² = 5

∴ The required equation of hyperbola is `x^2/4 - y^2/5` = 1.