The given curves are $\mathrm{x}=\mathrm{y}^2 \quad \ldots(i)$
and $x y=k \quad \ldots(ii)$
$\Rightarrow \mathrm{y}=\mathrm{k}^{1 / 3}$ from (i) $\mathrm{x}=\mathrm{y}^2=\left(\mathrm{k}^{1 / 3}\right)^2=\mathrm{k}^{2 / 3}$
Thus the point of intersection is $\left(\mathrm{k}^{2 / 3}, \mathrm{k}^{1 / 3}\right)$
Differentiating (i) w.r.t. x :
$1=2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}}$ at $\left(\mathrm{k}^{2 / 3}, \mathrm{k}^{1 / 3}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{k}^{1 / 3}} \Rightarrow$ Slope of tangent at $\left(\mathrm{k}^{2 / 3}, \mathrm{k}^{1 / 3}\right)$
Differentiating $x y=k$ w.r.t. $x$
$x \frac{d y}{d x}+y \cdot 1=0 \Rightarrow x \frac{d y}{d x}=-y \Rightarrow \frac{d y}{d x}=-\frac{y}{x}$
$\operatorname{At}\left(\mathrm{k}^{2 / 3}, \mathrm{k}^{1 / 3}\right), \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{k}^{1 / 3}}{\mathrm{k}^{2 / 3}}=\frac{-1}{\mathrm{k}^{1 / 3}}$
$\Rightarrow$ Slope of tangent at $\left(\mathrm{k}^{2 / 3}, \mathrm{k}^{1 / 3}\right)=\frac{-1}{\mathrm{k}^{1 / 3}}$
Now the curves cut at right angles if the product of slopes of tangents to two curves at $\left(\mathrm{k}^{2 / 3}, \mathrm{k}^{1 / 3}\right)$ is $-1$.
i.e., if $\left(\frac{1}{2 \mathrm{k}^{1 / 3}}\right)\left(\frac{-1}{\mathrm{k}^{1 / 3}}\right)=-1 \Rightarrow-1=2 \mathrm{k}^{2 / 3} \Rightarrow 8 \mathrm{k}^2=1$