Question

Find the equation of the hyperbola whose foci are ($\pm$ 5, 0), the transverse axis is of length 8.

Answer 1

Given length of transverse axis is $8=2a\Rightarrow a=4$

Also foci are $(\pm 5,0)$, so center of the ellipse is the mid point of foci, which is $(0,0)$

Also $ae=5\Rightarrow 4e=5\Rightarrow \frac{5}{4}$

Now using $b^2=a^2(e^2-1)=16(\frac{25}{16}-1)=9$

Hence the required hyperbola is, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1$