Is there a neat method for solving equations of the form: $$\frac {a^2}{x^2}-\frac {b^2}{(1-x)^2}=c^2$$ where $0<x<1$, by exploiting the symmetry in instead of expanding it as a fully-blown quartic?
Perhaps one could use the substitution $y=\frac 12-x$, but it doesn't seem to go far.
Additional Note: The original equation has $$\begin{align} a^2&=u^2\mu^2\\ b^2&=v^2(1-\mu)^2\\ c^2&=\frac {v^2-u^2}{\gamma^2}\end{align}$$ where $0<\mu <1$. Note sure if this additional information helps.
Above equation shown below:
$$\frac {a^2}{x^2}-\frac {b^2}{(1-x)^2}=c^2\tag1$$
Equation $(1)$ is actually a hyperbolic curve in disguise.
Taking, $x=(\cos p)^2$ we get $(1-x)=(\sin p)^2$.
For p=30 degree's we get, $(\sin p)^2 =\frac14$ &
$x=(\cos p)^2=\frac34$
Thus we get equation:
$$(4a)^2=(12b)^2+(3c)^2\tag2$$ Equation $(2)$ is satisfied at,
$(a,b,c)=[((5/4),(1/4),(4/3)]$
Hence solution for equation $(1)$:
$(a,b,c,x)=[((5/4),(1/4),(4/3),(3/4)]$