$\large\frac{x^2}{25}+\frac{y^2}{9}$$=1$
$e=\sqrt{1-\large\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1-\large\frac{9}{25}}$
$\Rightarrow \large\frac{4}{5}$
Foci of ellipse are $(\pm ae,0)$ i.e $(\pm 4,0)$
Let 'e' be the ecentricity of the required hyperbola and its equation be
$\large\frac{x^2}{A^2}-\frac{y^2}{B^2}$$=1$
The co-ordinate of foci are $(\pm Ae',0)$
$(A\times e')=4$
$A\times 3=4$
$A=\large\frac{4}{3}$
$B^2=A^2(e'^2-1)$
$B^2=\large\frac{16\times 8}{9}$
Substituting the value of A & B we get the equation of hyperbola
$\large\frac{x^2}{1}-\frac{y^2}{8}=\frac{16}{9}$
Hence (a) is the correct answer.