Prove that the two parabolas y² = 4ax and x²=4by intersect (other than  (0, 0)) at an angle of ${\tan }^{-1}\left[\frac{3{\mathrm{a}}^{1/3}{\mathrm{b}}^{1/3}}{2\left({\mathrm{a}}^{2/3}+{\mathrm{b}}^{2/3}\right)}\right]$)

${\mathrm{y}}^2=4\mathrm{ax}\to \left(1\right){\mathrm{x}}^2=4\mathrm{by}\to \left(2\right)$
Let P(x, y) be P.O.I of given parabolas
Then from (1) $\Rightarrow {\mathrm{y}}^4=16{\mathrm{a}}^2{\mathrm{x}}^2$
$\begin{array}{l}\Rightarrow {\mathrm{y}}^4=16{\mathrm{a}}^2\left(4\mathrm{by}\right)\Rightarrow {\mathrm{y}}^4=64{\mathrm{a}}^2\mathrm{by}\\ \Rightarrow {\mathrm{y}}^4-64{\mathrm{a}}^2\mathrm{by}=0\Rightarrow \mathrm{y}\left({\mathrm{y}}^3-64{\mathrm{a}}^2\mathrm{b}\right)=0\\ \Rightarrow {\mathrm{y}}^3-64{\mathrm{a}}^2\mathrm{b}=0(\because \mathrm{y}>0)\Rightarrow \mathrm{y}=(64{)}^{\frac{1}{3}}{\mathrm{a}}^{\frac{2}{3}}\cdot {\mathrm{b}}^{\frac{1}{3}}\\ \Rightarrow y=4a^{\frac{2}{3}}{b}^{\frac{1}{3}}\end{array}$
Then ${\mathrm{y}}^2=4\mathrm{ax}\Rightarrow \mathrm{x}=\frac{16{\mathrm{a}}^{\frac{4}{3}}{\mathrm{b}}^{\frac{2}{3}}}{4\mathrm{a}}$
point of intersection $\mathrm{P}\left(\frac{16{\mathrm{a}}^{4/3}{\mathrm{b}}^{2/3}}{4\mathrm{a}}\cdot 4{\mathrm{a}}^{2/3}{\mathrm{b}}^{1/3}\right)$
$\mathrm{P}\left(4{\mathrm{a}}^{1/3}{\mathrm{b}}^{2/3},4{\mathrm{a}}^{2/3}{\mathrm{b}}^{1/3}\right)$
Differentiating both sides of y² = 4ax w.r.t “x”
$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4\mathrm{a}}{2\mathrm{y}}=\frac{2\mathrm{a}}{\mathrm{y}} \\ Now {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{P}}=\frac{2\mathrm{a}}{4{\mathrm{a}}^{2/3}{\mathrm{b}}^{1/3}}=\frac{1}{2}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{1/3} \end{gathered}$$
If m₁ be the slope of (1) then
$$\begin{gathered} {\mathrm{m}}_1=\frac{1}{2}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{1/3} \\ \text{ similarly }{\mathrm{m}}_2=2{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\frac{1}{3}} \end{gathered}$$
If $\mathrm{\theta }$ is the acute angle between the tangents
$\begin{array}{l}then \tan \mathrm{\theta }=\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|\\ =\frac{2{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{1/3}-\frac{1}{2}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{1/3}}{1+{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{2/3}}\\ =\frac{3{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{1/3}}{2\left(1+{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{2/3}\right)}\\ =\frac{\frac{3{\mathrm{a}}^{1/3}}{{\mathrm{b}}^{1/3}}}{2\left(1+\frac{{\mathrm{a}}^{2/3}}{{\mathrm{b}}^{2/3}}\right)}\\ =\frac{3{\mathrm{a}}^{1/3}\cdot {\mathrm{b}}^{2/3}}{2{\mathrm{b}}^{1/3}\left({\mathrm{b}}^{2/3}+{\mathrm{a}}^{2/3}\right)}\\ \tan \mathrm{\theta }=\frac{3{\mathrm{a}}^{1/3}{\mathrm{b}}^{1/3}}{2\left({\mathrm{a}}^{2/3}+{\mathrm{b}}^{2/3}\right)}\\ \Rightarrow \mathrm{\theta }={\tan }^{-1}\left[\frac{3{\mathrm{a}}^{1/3}{\mathrm{b}}^{1/3}}{2\left({\mathrm{a}}^{2/3}+{\mathrm{b}}^{2/3}\right)}\right]\end{array}$