Question

The equation of the hyperbola whose foci are the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ and the eccentricity is $2$, is

Answer 1

If we take the question to mean that the foci of the ellipse are also the foci of an hyperbola$,$ then we have for the ellipse$,$

$a^2-c^2=b^2$

so$,$ $25-c^2=9$ and that means $c^2=16.$

This ellipse has its major axis on the $x-axis.$

For the hyperbola$,$ which must have its transverse axis on the $x-axis,$ the equation $c^2-a^2=b^2$ and $e=c/a=2.$

Only the $c$ value is he same as for the ellipse$;$ $c=4.$

Thus $4/a=2$ tells us that a $($for the hyperbola$)$ $=2.$

Therefore we compute $b^2=16-4=12.$

The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1;$ substituting give us $\frac{x^2-y^2}{12}=1.$