The foci of hyperbola coincide with the foci of the ellipse  $\frac{x^2}{25}+\frac{y^2}{9}=1$.then the equation of the hyperbola if its eccentricity is 2.

The given ellipse is  $\frac{x^2}{25}+\frac{y^2}{9}=1$
      
Comparing with    $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$a^2=25$      and  $b^2=9$
then   eccentricity  $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$
$\therefore$   Foci of ellipse are $(\pm ae,0)$  i.e,  $(\pm 4,0)$
So, the co-ordinates of foci of the hyperbola are  $(\pm 4,0)$
Let  $e\text{'}$  be the eccentricity of the required hyperbola and its equation be
$\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$      …… (1)
The co-ordinates of foci are  $(\pm Ae\text{'},0)$
   $Ae\text{'}=4\Rightarrow A\times 2=4\Rightarrow A=2$
Also   $B^2=A^2(e{\text{'}}^2-1)$                 (given  $e\text{'}=2$)
  $=4(4-1)=12$
Substituting the values of A and B in (1), we get
  $\frac{x^2}{4}-\frac{y^2}{12}=1$
or  $3x^2-y^2-12=0$
which is required hyperbola.