Question

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

whose distance between foci is 10 and eccentricity `5/2`

Answer 1

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

Given, eccentricity (e) = `5/2`

Distance between foci = 2ae

Given, distance between foci = 10

∴ 2ae = 10

∴ ae = `10/2` = 5

∴ `"a"(5/2)` = 5

∴ a = 2

∴ a² = 4

Now, b² = a²(e² – 1)

∴ b² = `4[(5/2)^2 - 1]`

= `4(25/4 - 1)`

= `4(21/4)`

∴ b² = 21

∴ The required equation of hyperbola is `x^2/4 - y^2/21` = 1.