Question

The distance between the foci of a hyperbola is 16 and its eccentricity is `sqrt(2)`. Its equation is ______.

Options

• x² – y² = 32

• `x^2/4 - y^2/9` = 1

• 2x² – 3y² = 7

• None of these

Answer 1

The distance between the foci of a hyperbola is 16 and its eccentricity is `sqrt(2)`. Its equation is x² – y² = 32.

Explanation:

We know that the distance between the foci = 2ae

∴ 2ae = 16

⇒ ae = 8

Given that e = `sqrt(2)`

∴ `sqrt(2)a` = 8

⇒ `a = 4sqrt(2)`

Now b² = a² (e²  – 1)

⇒ b² = 32(2 – 1)

⇒ b² = 32

So, the equation of the hyperbola is `x^2/a^2 - y^2/b^2` = 1

⇒ `x^2/32 - y^2/32` = 1

⇒ x² – y² = 32