Question

Find the equation of the hyperbola with eccentricity `3/2` and foci at (± 2, 0).

Answer 1

Given that e = `3/2` and foci at (± 2, 0)

We know that foci = (± ae, 0)

∴ ae = 2

⇒ `a xx 3/2` = 2

⇒ `a = 4/3`

⇒ `a^2 = 16/9`

We know that b² = a²(e² – 1)

⇒ `b^2 = 16/9(9/4 - 1)`

= `16/9 xx 5/4`

= `20/9`

So, the equation of the hyperbola is `x^2/(16/9) - y^2/(20/9)` = 1

⇒ `(9x^2)/16 - (9y^2)/20` = 1

⇒ `x^2/4 - y^2/5 = 4/9`

Hence, the required equation is `x^2/4 - y^2/5 = 4/9`.