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

1 Answer
Ratnaker Mehta
Sep 14, 2017

x^2/48-y^2/16=1.x248−y216=1.

Explanation:

Recall that, for the Hyperbola S : x^2/a^2-y^2/b^2=1,S:x2a2−y2b2=1,

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

Here, e > 1,e>1, the Eccentricity, is given by, b^2=a^2(e^2-1).b2=a2(e2−1).

:. 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.

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