$x^2-y^2=4$ ...(i)
$x^2+y^2=4 \sqrt{2}$ ...(ii)
Adding both equations,
$\begin{aligned} & x^2=2(1+\sqrt{2}) \Rightarrow x=\sqrt{2} \sqrt{\sqrt{2}+1} \\ & \therefore \quad y^2=2(\sqrt{2}-1) \Rightarrow y=\sqrt{2} \sqrt{\sqrt{2}-1}\end{aligned}$
Differentiating both (i) and (ii)
$\begin{aligned} & 2 x-2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{x}{y}=m_1 \\ & \text { and } 2 x+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-x}{y}=m_2 \\ & \tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\left|\frac{\frac{x}{y}+\frac{x}{y}}{1-\frac{x^2}{y^2}}\right| \\ & \quad=\left|2 \frac{x}{y} \times \frac{y^2}{y^2-x^2}\right|=\left|\frac{2 x y}{y^2-x^2}\right|=\left|\frac{4}{-4}\right|=1 . \\ & \therefore \quad \theta=\frac{\pi}{4} .\end{aligned}$