Angle of intersection between the curves ${\mathrm{x}}^2+{\mathrm{y}}^2=\sqrt{2}{\mathrm{a}}^2$ and ${\mathrm{x}}^2-{\mathrm{y}}^2={\mathrm{a}}^2$ is

$\begin{array}{rcl}{\mathrm{x}}^2+{\mathrm{y}}^2& =& \sqrt{2}{\mathrm{a}}^2 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{x}}{\mathrm{y}}={\mathrm{m}}_1\\ {\mathrm{x}}^2-{\mathrm{y}}^2& =& {\mathrm{a}}^2 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{\mathrm{y}}={\mathrm{m}}_2\\ \mathrm{tan\theta }& =& \left|\frac{2{\displaystyle \frac{\mathrm{x}}{\mathrm{y}}}}{1-{\displaystyle \frac{{\mathrm{x}}^2}{{\mathrm{y}}^2}}}\right|=\left|\frac{2\mathrm{xy}}{{\mathrm{y}}^2-{\mathrm{x}}^2}\right|=\left|\frac{2\mathrm{xy}}{{\mathrm{a}}^2}\right|\end{array}$

adding two equations $2{\mathrm{x}}^2=\left(\sqrt{2}+1\right){\mathrm{a}}^2$

                                        $2{\mathrm{y}}^2=\left(\sqrt{2}-1\right){\mathrm{a}}^2$

$\mathrm{tan\theta }=\left|\frac{2{\displaystyle \frac{{\mathrm{a}}^2}{2}}}{{\mathrm{a}}^2}\right|=1 \Rightarrow \mathrm{\theta }=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$