Let the equation of hyperbola be
$$
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
$$
Given equation of ellipse is
$$
\frac{x^2}{(13)^2}+\frac{y^2}{(5)^2}=1
$$
Here,
$$
\begin{aligned}
& \text { Here, } \quad a=13, b=5 \\
& \therefore \quad e=\sqrt{1-\frac{b^2}{a^2}} \\
& \therefore \quad=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13} \\
& \therefore \text { Focus }( \pm a e, 0)=\left( \pm 13 \times \frac{12}{13}, 0\right) \\
& \quad=( \pm 12,0)
\end{aligned}
$$
Since, Eq. (i) passes through $( \pm 12,0)$.
$$
\begin{array}{rrrl}
\therefore & \frac{144}{a^2}-\frac{0}{b^2}=1 \\
\Rightarrow & a^2=144 \\
\Rightarrow & a= \pm 12
\end{array}
$$
Now eccentricity of hyperbola
$$
\begin{aligned}
e^{\prime} & =\sqrt{1+\frac{b^2}{a^2}} \\
& =\sqrt{1+\frac{b^2}{144}}
\end{aligned}
$$
According to the equation,
$$
\begin{array}{rlrl}
& e^{\prime}=1 \\
\Rightarrow & & \frac{12}{13} \times \sqrt{1+\frac{b^2}{144}} & =1 \\
\Rightarrow & & \sqrt{1+\frac{b^2}{144}} & =\frac{13}{12} \\
\Rightarrow & & 1+\frac{b^2}{144} & =\frac{169}{144} \\
\Rightarrow & & \frac{b^2}{144} & =\frac{169}{144}-1 \\
\Rightarrow & & \frac{b^2}{144} & =\frac{25}{144} \\
b^2 & =25
\end{array}
$$
$\therefore$ Equation of hyperbola is
$$
\frac{x^2}{144}-\frac{y^2}{25}=1
$$