Question

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

Answer 1

Since the foci of the given hyperbola are of the form $(\pm c,0),$ it is a horizontal hyperbola.

Let the required equation be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$

Length of its transverse axis = 2a.

$\therefore 2a=8\iff a=4\iff a^2=16.$

Let its foci be $(\pm c,0).$

Then, $c=5$ $[\because fociare(\pm 5,0\left)\right].$

$\therefore b^2=(c^2-a^2)=(5^2-4^2)=(25-16)=9.$

Thus, $a^2=16$ and $b^2=9.$

Hence, the required equation is $\frac{x^2}{16}-\frac{y^2}{9}=1.$