Question

How do you write the equation of the hyperbola given Foci: (-8,0),(8,0) and vertices (-4sqrt3,0), (4sqrt3,0)?

Answer 1

`x^2/48-y^2/16=1.`

Explanation:

Recall that, for the Hyperbola `S : x^2/a^2-y^2/b^2=1,`

Focii and Vertices are, `(+-ae,0), and, (+-a,0),` resp.

Here, `e > 1,` the Eccentricity, is given by, `b^2=a^2(e^2-1).`

`:. ae=8, and, a=4sqrt3.`

Now, `b^2=a^2(e^2-1)=a^2e^2-a^2=8^2-(4sqrt3)^2=16.`

Hence, the desired eqn. of the Hyperbola is, `x^2/48-y^2/16=1.`