If the curves \(2 x=y^{2}\) and \(2 x y=K\) intersect perpendicularly, then the value of \(K^{2}\) is

Given curves are
\(2 x=y^{2}...(i)\)
and \(2 x y=K...(ii)\)
On solving both Eqs. (i) and (ii), we get
\(x=\frac{K^{2 / 3}}{2} \text { and } y=K^{1 / 3}\)
\(\therefore\) Intersecting point of both curves is \(\left(\frac{K^{2 / 3}}{2}, K^{1 / 3}\right)\)
Now differentiate Eq. (i) w.r.t. \(x\), we get
\(\begin{aligned}
2 &=2 y \frac{d y}{d x} \\
\Rightarrow \quad \frac{d y}{d x} &=\frac{1}{y}
\end{aligned}\)
\(\therefore\) Slope of tangent at \(\left(\frac{K^{2 / 3}}{2}, K^{1 / 3}\right)=\frac{1}{K^{1 / 3}}\)
And differentiate Eq. (ii) w.r.t. \(x\), we get
\(2\left(x \frac{d y}{d x}+y\right)=0 \Rightarrow \frac{d y}{d x}=\frac{-y}{x}\)
\(\therefore\) Slope of tangent at \(\left(\frac{K^{2 / 3}}{2}, K^{1 / 3}\right)\)
\(=\frac{-K^{1 / 3}}{K^{2 / 3} / 2}=-2 K^{-1 / 3}\)
Since, both curves intersect perpendicularly
\(\begin{aligned}
&\therefore \quad \frac{1}{K^{1 / 3}} \times\left(-2 K^{-1 / 3}\right)=-1 \\
&\Rightarrow \quad-2 K^{-2 / 3}=-1 \Rightarrow K^{2 / 3}=2 \Rightarrow K^{2}=8
\end{aligned}\)