The condition that the two curves $\mathrm{x}={\mathrm{y}}^2$, $xy=k$ cut orthogonally is

$\begin{array}{l}\Rightarrow \frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(y^2\right)\Rightarrow 1=2y\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{1}{2y}=m_1\\ \frac{d}{dx}\left(xy\right)=\frac{d}{dx}\left(k\right)\Rightarrow y+x\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{-y}{x}=m_2\\ Two curves are cuts orthogonally,then\\ \therefore m_1\cdot m_2=-1\\ \left(\frac{1}{2y}\right)\cdot \left(\frac{-y}{x}\right)=-1\Rightarrow x=\frac{1}{2}\\ x=y^2\Rightarrow y=\frac{1}{\sqrt{2}}\therefore xy=k\\ \Rightarrow \frac{1}{2}\cdot \frac{1}{\sqrt{2}}=k\Rightarrow 8k^2=1\end{array}$