The equation of ellipse is given as
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{7}=1$
Eccentricity is given by
$e=\sqrt{1-\frac{7}{a^{2}}}$
Therefore, foci of ellipse are $(\pm \mathrm{ae}, 0) \mathrm{ie}$,
$\left(\pm a \sqrt{1-\frac{7}{a^{2}}}, 0\right)$
Now, the equation of given hyper bola is
$\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25} \Rightarrow \frac{x^{2}}{\frac{144}{25}}-\frac{y^{2}}{\frac{81}{25}}=1$
So, a $=\frac{12}{5}$ and $b=\frac{9}{5}$
$\therefore \quad e^{\prime}=\sqrt{1+\frac{81 / 25}{144 / 25}}=\sqrt{\frac{144+81}{144}}=\sqrt{\frac{225}{144}}$
$\quad=\frac{15}{12}$
Foci of hyper bola are $\left(\pm \frac{12}{5}, \frac{15}{12}, 0\right)$ ie, $(\pm 3,0)$.
Since these foci coincides
$\Rightarrow 3=a \sqrt{1-\frac{7}{a^{2}}}$
$\Rightarrow \frac{3}{\mathrm{a}}=\sqrt{1-\frac{7}{\mathrm{a}^{2}}}$
$\Rightarrow \frac{9}{\mathrm{a}^{2}}=1-\frac{7}{\mathrm{a}^{2}}$
$\Rightarrow \frac{16}{a^{2}}=1 \Rightarrow a=4$