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Question

Find equation of a hyperbola whose vertices are (±7,0) and the eccentricity is 43.

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Solution

Since vertices are (±7,0)
a=7
And e=1+b2a243=1+b2a2
So, b=737
So, equation of hyperbola be x2a2y2b2=1
So, equation is x249y2342=1
7x29y2=343

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