Given equation is $a x^2+b y^2=15$
$$
\frac{x^2}{\left(\sqrt{\frac{15}{a}}\right)^2}+\frac{y^2}{\left(\sqrt{\frac{15}{b}}\right)^2}=1
$$
Now, $a^{\prime}=\sqrt{\frac{15}{a}}, b^{\prime}=\sqrt{\frac{15}{b}}$
According to question,
$$
\begin{aligned}
& 2 d e=2 \Rightarrow a^{\prime} e=1 \\
& \frac{2 a^{\prime}}{e}=5 \Rightarrow \frac{a^{\prime}}{e}=\frac{5}{2}
\end{aligned}
$$
Multiply (ii) and (iii),
$$
a^{\prime} e \times \frac{a^{\prime}}{e}=1 \times \frac{5}{2}
$$
$$
a^{\prime 2}=\frac{5}{2} a^{\prime}= \pm \frac{\sqrt{5}}{\sqrt{2}}
$$
from (i)
$$
a^{\prime} e=1
$$
$$
\begin{aligned}
& e= \pm \frac{\sqrt{2}}{\sqrt{5}} \\
& b^{\prime 2}=a^{\prime 2}=\left(1-e^2\right) \\
& b^{\prime 2}=\frac{5}{2}\left(1-\frac{2}{5}\right)=\frac{5}{2} \times \frac{3}{5}=\frac{3}{2} \\
& b^{\prime}= \pm \frac{\sqrt{3}}{\sqrt{2}}
\end{aligned}
$$
Now, compare $a^{\prime} \& b^{\prime}$ values,
$$
\frac{\sqrt{15}}{\sqrt{a}}= \pm \frac{\sqrt{5}}{\sqrt{2}}
$$
square both sides,
$$
\frac{15}{a}=\frac{5}{2} \Rightarrow a=6 \text {. }
$$
Similarly, $\frac{\sqrt{15}}{\sqrt{b}}= \pm \frac{\sqrt{3}}{\sqrt{2}}$
$$
\begin{aligned}
& \frac{15}{b}=\frac{3}{2} \Rightarrow b=10 \\
& \text { So, } a+b=10+6=16
\end{aligned}
$$