Question

Answer the following:

Find the equation of the hyperbola in the standard form if eccentricity is `3/2` and distance between foci is 12.

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 foci = 2ae

Given, distance between foci = 12

∴ 2ae = 12

∴ `2"a"(3/2)` = 12

∴ 3a = 12

∴ a = `12/3` = 4

∴ a² = 16

Now, b² = a²(e² – 1)

∴ b² = `16[(3/2)^2 - 1]`

= `16(9/4 - 1)`

= `16(5/4)`

∴ b² = 20

∴ The required equation of hyperbola is `x^2/16 - y^2/20` = 1.