Question

Find equation of a hyperbola whose vertices are $(\pm 7,0)$ and the eccentricity is $\frac{4}{3}$.

Answer 1

Since vertices are $(\pm 7,0)$

$a=7$

And $e=\sqrt{1+\frac{b^2}{a^2}}⟹\frac{4}{3}=\sqrt{1+\frac{b^2}{a^2}}$

So, $b=\frac{7}{3}\sqrt{7}$

So, equation of hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

So, equation is $\frac{x^2}{49}-\frac{y^2}{342}=1$

$⟹7x^2-9y^2=343$