Given two points $\mathrm{P}(\theta)$ and $\mathrm{Q}\left(\frac{\pi}{2}+\theta\right)$ are on the ellipse. Hence,
$$
\mathrm{P}=(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta), \mathrm{Q}=(-\mathrm{a} \sin \theta, \mathrm{b} \cos \theta)
$$
now mid point of $P Q$
$$
\begin{aligned}
& =\left(\frac{\mathrm{a} \cos \theta-\mathrm{a} \sin \theta}{2}, \frac{\mathrm{b} \sin \theta+\mathrm{b} \cos \theta}{2}\right) \\
& \Rightarrow \frac{\mathrm{x}}{\mathrm{a}}=\frac{\cos \theta-\sin \theta}{2} \text { and } \frac{4}{\mathrm{~b}}=\frac{\sin \theta+\cos \theta}{2}
\end{aligned}
$$
On squaring and adding we get
$$
\begin{aligned}
& \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=\frac{1}{2}\left(\text { Given as } \frac{\mathrm{x}^2}{\alpha^2}+\frac{\mathrm{y}^2}{\beta^2}=1\right) \\
& \Rightarrow \frac{\mathrm{x}^2}{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)^2}+\frac{\mathrm{y}^2}{\left(\frac{\mathrm{b}}{\sqrt{2}}\right)^2}=1 \\
& \Rightarrow \alpha=\frac{9}{\sqrt{2}}, \beta=\frac{\mathrm{b}}{\sqrt{2}} \\
& \therefore \frac{\mathrm{a}+\mathrm{b}}{\alpha+\beta}=\frac{\mathrm{a}+\mathrm{b}}{\frac{1}{\sqrt{2}}(\mathrm{a}+\mathrm{b})}=\sqrt{2}
\end{aligned}
$$