An angle between the curves $x^2-y^2=4$ and $x^2+y^2=4 \sqrt{2}$ is

Equation of given curves, $x^2-y^2=4$ and $x^2+y^2=4 \sqrt{2}$ Let $\left(x_0, y_0\right)$ be point of intersection. $\therefore$ Angle between curves $=$ Angle between tangent at point of intersection. Now, $x^2-y^2=4$ Differentiate w.r.t. ' $x$ ' $$ 2 x-2 y \frac{d y}{d x}=0 \quad \frac{d y}{d x}=\frac{x}{y} $$ At point $\left(x_0, y_0\right)$ $$ m_1=\frac{x_0}{y_0} \text { and curve } x^2+y^2=4 \sqrt{2} $$ Differentiate w.r.t. ' $x$ ' $$ 2 x+2 y \frac{d y}{d x}=0 \quad \frac{d y}{d x}=-\frac{x}{y} $$ At point $\left(x_0, y_0\right)$ $$ m_2=-\frac{x_0}{y_0} $$ Angle between tangents $\theta=\tan ^{-1}\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$ $$ \begin{aligned} & \theta=\tan ^{-1}\left|\frac{\frac{x_0}{y_0}+\frac{x_0}{y_0}}{1+\frac{x_0}{y_0}\left(-\frac{x_0}{y_0}\right)}\right| \Rightarrow \theta=\tan ^{-1}\left|\frac{2 \frac{x_0}{y_0}}{\frac{y_0^2-x_0^2}{y_0^2}}\right| \\ & \Rightarrow \quad \theta=\tan ^{-1}\left(\frac{2 x_0 y_0}{y_0^2-x_0^2}\right)=\tan ^{-1}\left(\frac{2 x_0 y_0}{4}\right) \end{aligned} $$ [given, $y^2-x^2=4$ satisfied point $\left(x_0, y_0\right)$ ] $$ =\tan ^{-1}\left(\frac{x_0 y_0}{2}\right) $$ Point $\left(x_0, y_0\right)$ lie on curves, so $$ \begin{array}{rlrl} & & x_0^2-y_0^2 & =4 \text { and } x_0^2+y_0^2=4 \sqrt{2} \\ \Rightarrow & & x_0^2 & =2(1+\sqrt{2}) \text { and } y_0^2=2(\sqrt{2}-1) \\ & \text { So, } \quad & x_0^2 y_0^2 & =4(\sqrt{2}+1)(\sqrt{2}-1)=4(2-1)=4 \\ x_0 y_0 & =2 \end{array} $$ Put in Eq. (i), $$ \theta=\tan ^{-1}\left(\frac{2}{2}\right) \quad \theta=\tan ^{-1}(1) \Rightarrow \theta=\frac{\pi}{4} $$