Given equation of pair of lines is
$\begin{aligned}
& \mathrm{a} x^2+(2 \mathrm{a}+1) x y+2 y^2=0 \\
& \mathrm{~A}=\mathrm{a}, \mathrm{H}=\frac{2 \mathrm{a}+1}{2}, \mathrm{~B}=2
\end{aligned}$
Given condition,
$\begin{array}{ll}
& \mathrm{m}_1=\frac{1}{\mathrm{~m}_2} \\
\therefore \quad & \mathrm{m}_1 \cdot \mathrm{m}_2=1 \\
& \text { Product of slopes }=\frac{\mathrm{A}}{\mathrm{B}}=\frac{\mathrm{a}}{2} \\
\therefore \quad & \mathrm{m}_1 \cdot \mathrm{m}_2=1=\frac{\mathrm{a}}{2} \\
\therefore \quad & \mathrm{a}=2 \\
& \text { Also, sum of slopes }=\frac{-2 \mathrm{H}}{\mathrm{B}}=-\left(\frac{2 \mathrm{a}+1}{2}\right)=\frac{-5}{2} \\
& \text { Using }\left(\mathrm{m}_1+\mathrm{m}_2\right)^2=\mathrm{m}_1^2+\mathrm{m}_2^2+2 \mathrm{~m}_1 \mathrm{~m}_2 \\
& \left(\frac{-5}{2}\right)^2=\mathrm{m}_1^2+\mathrm{m}_2^2+2 \times 1 \\
\therefore \quad & \mathrm{m}_1^2+\mathrm{m}_2^2=\frac{25}{4}-2 \\
\therefore \quad & \mathrm{m}_1^2+\mathrm{m}_2^2=\frac{17}{4}
\end{array}$