Centre is
$$
\frac{\left(x-\frac{\alpha}{2}\right)^{2}}{a^{2}}+\frac{(y-\beta)^{2}}{b^{2}}=1
$$
pass through $(0,0)(0,2) \&(1,0)$
Distance between $\begin{aligned} F_{1} \& F_{2} \Rightarrow \alpha &=2 \mathrm{ae} \\ \alpha^{2} &=4 \mathrm{a}^{2} \mathrm{e}^{2} \end{aligned}$
Pass through $(0,0)$
$$
\begin{array}{l}
\Rightarrow \frac{\alpha^{2}}{4 \mathrm{a}^{2}}+\frac{\beta^{2}}{\mathrm{~b}^{2}}=1 \\
\frac{\alpha^{2}}{4 \mathrm{a}^{2}}=\mathrm{e}^{2}
\end{array}
$$
Pass through $(0,2)$
$$
\frac{\alpha^{2}}{4 \mathrm{a}^{2}}+\frac{(2-\beta)^{2}}{\mathrm{~b}^{2}}=1
$$
from these two $\Rightarrow \beta=1$
Use properties $\mathrm{PF}_{1}+\mathrm{PF}_{2}=2 \mathrm{a}$
$$
\begin{array}{l}
\mathrm{F}_{1}(0,1) \quad \mathrm{F}_{2}(\alpha, 1) \quad \mathrm{P}(0,2) \\
1+\sqrt{\alpha^{2}+1}=2 \mathrm{a} \\
\mathrm{F}_{1}(0,1) \quad \mathrm{F}_{2}(\alpha, 1) \quad \mathrm{P}(1,0) \\
\sqrt{1+1}+\sqrt{(\alpha-1)^{2}+1}=2 \mathrm{a}
\end{array}
$$
From these two
$$
\alpha=-2 \alpha+2 \sqrt{2} \mathrm{a}
$$
put $\alpha$ in any of above two equations
$$
\begin{array}{l}
a=\frac{\sqrt{2}+1}{2} \Rightarrow 2 a=\sqrt{2}+1 \\
\alpha=1 \\
\alpha=2 \mathrm{ae} \\
\alpha=1,2 \mathrm{a}=\sqrt{2}+1 \\
\text { find } \mathrm{e}=\sqrt{2}
\end{array}
$$