Solve $\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$ for $y$ I have the equation:
$\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$,
where $k$ and $c$ are arbitrary constants.
How do I go about simplifying this and solving for $y$ in terms of $x$, excluding the obvious solution $y=-x$

$$\frac{ky^2}{2}+cy=\frac{kx^2}{2}-cx \iff y^2+\frac{2cy}{k}=\frac2k \left (\frac{kx^2}{2}-ck \right ) \iff y^2+\frac{2cy}{k}+\frac{c^2}{k^2}=\frac2k \left (\frac{kx^2}{2}-ck \right )+\frac{c^2}{k^2} \iff \left (y+\frac{c}{k}\right )^2 =\frac{(c-kx)^2}{k^2}$$
I think you can continue from here.
