Show that the common tangent to the parabola y²=4ax and x²= 4by is xa^1/3+yb^1/3 +a^2/3b^2/3 = 0

Given ${\mathrm{y}}^2=4\mathrm{ax}\to \left(1\right),{\mathrm{x}}^2=4\mathrm{by}\to \left(2\right)$ be parabolas
Let equations of tangent to (1) is
$\mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\to \left(3\right)$
I f (3) is also tangent to (2) then substitute y value of (3) in (2)
$\begin{array}{l}\Rightarrow {\mathrm{x}}^2=4\mathrm{b}\left(\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\right)\Rightarrow {\mathrm{x}}^2-4\mathrm{b}\left(\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\right)=0\\ \Rightarrow {\mathrm{mx}}^2-4\mathrm{b}\left({\mathrm{m}}^2\mathrm{x}+\mathrm{a}\right)=0\\ \Rightarrow {\mathrm{mx}}^2-4{\mathrm{bm}}^2\mathrm{x}-4\mathrm{ab}=0\end{array}$
It is in the form Q.E of x terms

 roots are equal then $∆$ = 0
$\begin{array}{l}\Rightarrow 16{\mathrm{b}}^2{\mathrm{m}}^4+4\left(\mathrm{m}\right)\left(4\mathrm{ab}\right)=0\\ \Rightarrow 16{\mathrm{b}}^2{\mathrm{m}}^4+16\mathrm{mab}=0\Rightarrow 16\mathrm{b}\left({\mathrm{bm}}^4+\mathrm{am}\right)=0\\ \Rightarrow {\mathrm{bm}}^4+\mathrm{am}=0\Rightarrow \mathrm{m}\left({\mathrm{bm}}^3+\mathrm{a}\right)=0(\because \mathrm{m}\ne 0)\\ \because {\mathrm{bm}}^3+\mathrm{a}=0\Rightarrow \mathrm{m}=-\frac{{\mathrm{a}}^{\frac{1}{3}}}{{\mathrm{b}}^{\frac{1}{3}}}\end{array}$
$\because \text{ equation of tangent to }\left(2\right)\text{ is }$
$\begin{array}{l}\Rightarrow \mathrm{y}=\left(-{\mathrm{a}}^{1/3}/{\mathrm{b}}^{1/3}\right)\mathrm{x}+\frac{\mathrm{a}}{\left(-{\mathrm{a}}^{1/3}/{\mathrm{b}}^{1/3}\right)}\\ \Rightarrow \mathrm{y}={\left(-\frac{\mathrm{a}}{\mathrm{b}}\right)}^{1/3}\mathrm{x}+\frac{\mathrm{a}}{{\left(-\frac{\mathrm{a}}{\mathrm{b}}\right)}^{1/3}}\\ \Rightarrow \mathrm{x}\cdot {\mathrm{a}}^{1/3}+\mathrm{y}\cdot {\mathrm{b}}^{1/3}+{\mathrm{a}}^{2/2}{\mathrm{b}}^{2/3}=0\end{array}$