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

1 Answer

Sep 14, 2017

#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.#

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