Given $k y^2=2(y-x)$ .........(i)

$$ 2 y^2=k x $$ .........(ii)

Point of intersection of (i) and (ii)

$$ \begin{aligned} & k y^2=2\left(y-\frac{2 y^2}{k}\right) \\\\ & \Rightarrow y=0, k y=2\left(1-\frac{2 y}{k}\right) \end{aligned} $$

$\begin{aligned} & k y+\frac{4 y}{k}=2 \\\\ & y=\frac{2}{k+\frac{4}{k}}=\frac{2 k}{k^2+4}\end{aligned}$

$\begin{aligned} & A=\int\limits_6^{\frac{2 k}{k^2+4}}\left(\left(y-\frac{k y^2}{2}\right)-\frac{2 y^2}{k}\right) d y \\\\ & A=\left[\frac{y^2}{2}-\left(\frac{k}{2}+\frac{2}{k}\right) \frac{y^3}{3}\right]_0^{\frac{2 k}{k^2+4}}\end{aligned}$

$\begin{aligned} & =\left(\frac{2 k}{k^2+4}\right)^2\left[\frac{1}{2}-\frac{k^2+4}{2 k}\left(\frac{1}{3}\right)\left(\frac{2 k}{k^2+4}\right)\right] \\\\ & =\frac{1}{6} \times 4 \times\left(\frac{1}{k+\frac{4}{k}}\right)^2\end{aligned}$

$\begin{aligned} & \text { A.M. } \geq \text { G.M. } \\\\ & \frac{\left(k+\frac{4}{k}\right)}{2} \geq 2 \\\\ & k+\frac{4}{k} \geq 4\end{aligned}$

$\therefore$ Area is maximum when $k=\frac{4}{k}$

$$ \begin{aligned} & \therefore k^2=4 \\\\ & k= \pm 2 \\\\ & k_1=2, k_2=-2 \\\\ & \therefore k_1^2+k_2^2=(+2)^2+(-2)^2 \\\\ & =4+4 \\\\ & =08 \end{aligned} $$