The sum of squares of all possible values of $$k$$, for which area of the region bounded by the parabolas $$2y^2 = kx$$ and $$ky^2 = 2(y - x)$$ is maximum, is equal to:

First rewrite the equations of the parabolas: Parabola 1 is $$2y^2 = kx\Rightarrow x=\frac{2y^2}{k}$$ and Parabola 2 is $$ky^2=2y-2x\Rightarrow x=y-\frac{ky^2}{2}\,. $$ Equating these expressions gives

$$\frac{2y^2}{k}=y-\frac{ky^2}{2}\quad\Longrightarrow\quad\frac{2y^2}{k}+\frac{ky^2}{2}=y\quad\Longrightarrow\quad y^2\Bigl(\frac{2}{k}+\frac{k}{2}\Bigr)=y\,. $$

For $$y\neq0$$ this yields $$y\Bigl(\frac{4+k^2}{2k}\Bigr)=1$$ and hence $$y=\frac{2k}{4+k^2}\,. $$ Thus the curves intersect at $$y=0$$ and $$y=\frac{2k}{4+k^2}\,. $$

The area between the curves, integrating with respect to $$y$$ from $$0$$ to $$y_0=\frac{2k}{4+k^2}$$, is

$$ A=\int_{0}^{y_0}\Bigl[\Bigl(y-\frac{ky^2}{2}\Bigr)-\frac{2y^2}{k}\Bigr]\,dy =\int_{0}^{y_0}\Bigl[y-y^2\Bigl(\frac{k}{2}+\frac{2}{k}\Bigr)\Bigr]\,dy =\int_{0}^{y_0}\Bigl[y-y^2\frac{k^2+4}{2k}\Bigr]\,dy. $$

Let $$\lambda=\frac{k^2+4}{2k}$$. Then

$$ A=\left[\frac{y^2}{2}-\frac{\lambda y^3}{3}\right]_0^{y_0} =\frac{y_0^2}{2}-\frac{\lambda y_0^3}{3}. $$

Since $$y_0\lambda=1$$ it follows that

$$ A=\frac{1}{2\lambda^2}-\frac{1}{3\lambda^2}=\frac{1}{6\lambda^2} =\frac{1}{6}\cdot\frac{4k^2}{(k^2+4)^2} =\frac{2k^2}{3(k^2+4)^2}\,. $$

To maximize this area, set $$u=k^2>0$$ so that $$A=\frac{2u}{3(u+4)^2}\,. $$ Differentiating with respect to $$u$$ gives

$$ \frac{dA}{du} =\frac{2}{3}\cdot\frac{(u+4)^2-2u(u+4)}{(u+4)^4} =\frac{2}{3}\cdot\frac{4-u}{(u+4)^3}. $$

Setting $$\frac{dA}{du}=0$$ yields $$u=4$$, hence $$k^2=4$$ and $$k=\pm2$$. For $$u<4$$ the derivative is positive and for $$u>4$$ it is negative, confirming a maximum at $$u=4\,. $$

Checking both values, when $$k=2$$ one has $$y_0=\tfrac{4}{8}=\tfrac12>0$$, and for $$k=-2$$ the limits reverse from $$-\tfrac12$$ to $$0$$, still producing the same positive area. Thus both values are valid.

The sum of squares of these values is $$2^2+(-2)^2=4+4=8\,$$