The angle between the curves $x^2=8 y$ and $x y=8$ is

Given equation of curves are $$ \begin{aligned} & x^2=8 y \\ & x y=8 \end{aligned} $$ On substituting the value of $y$ from Eq. (i) in Eq. (ii), we get $$ \begin{aligned} x\left(\frac{x^2}{8}\right) & =8 \\ x^3 & =64 \\ x & =4 \end{aligned} $$ $$ \begin{array}{ll} \Rightarrow & x^3=64 \\ \Rightarrow & x=4 \end{array} $$ On substituting $x=4$ in Eq. (ii), we get $y=2$ Thus, the point of intersection of given curves is $(4,2)$. Now, let $m_1$ be the slope of tangent to the curve (i) at point $(4,2)$ and $m_2$ be the slope of tangent to the curve (ii) at point $(4,2)$ Then, $m_1=\frac{4}{4}=1$ $$ \left[\because x^2=8 y \Rightarrow 2 x=8 \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{x}{4}\right] $$ and $m_2=\frac{-2}{4}=-\frac{1}{2}$ $$ \left[\because x y=8 \Rightarrow x \frac{d y}{d x}+y=0 \Rightarrow \frac{d y}{d x}=\frac{-y}{x}\right] $$ Now, the angle $\theta$ between the curves is given by $$ \begin{aligned} & \tan \theta=\left(\frac{m_2-m_1}{1+m_1 m_2}\right) \\ & =\left(\frac{\frac{-1}{2}-1}{1-\frac{1}{2}}\right)=\left(\frac{\frac{-3}{2}}{\frac{1}{2}}\right) \\ & \Rightarrow \quad \tan \theta=-3 \\ & \Rightarrow \quad \theta=\tan ^{-1}(-3) \\ & \end{aligned} $$