Question

Equation of the hyperbola with eccentricty `3/2` and foci at (± 2, 0) is ______.

Options

• `x^2/4 - y^2/5 = 4/9`

• `x^2/9 - y^2/9 = 4/9`

• `x^2/4 - y^2/9` = 1

• None of these

Answer 1

Equation of the hyperbola with eccentricty `3/2` and foci at (± 2, 0) is `x^2/4 - y^2/5 = 4/9`.

Explanation:

Given that e = `3/2`

And foci = (± ae, 0) = (± 2, 0)

∴ ae = 2

`a xx 3/2` = 2

⇒ `a = 4/3`

Now we know that b² = a²(e² – 1)

b² = `16/9(9/4 - 1)`

b² = `16/9 xx 5/4`

b² = `20/9`

So, the equation of the hyperbola is `x^2/(4/3)^2 - y^2/(20/9)` = 1

⇒ `(9x^2)/16 - (9y^2)/20` = 1

⇒ `x^2/16 - y^2/20 = 1/9`

⇒ `x^2/4 - y^2/5 = 4/9`