Question

Find the angle between the prabolas $y^2=4ax$ and $x^2=4by$ at their point of intersection other than origin.

Answer 1

The equations of two parabolas are $y^2=4ax$ and $x^2=4by$.
Now, $x^2=4by\Rightarrow y=\frac{x^2}{4b}$
Substituting this value of $y$ in $y^2=4ax$, we get
${\left(\frac{x^2}{4b}\right)}^2=4ax$
$\Rightarrow x^4-64a{b}^{2}x=0$
$\Rightarrow x(x^3-64a{b}^2)=0$
$\Rightarrow x=0$ or $x^3=64a{b}^2$
$\Rightarrow x=0$ or $x=4a^{1/3}{b}^{2/3}$
So, $y=\frac{x^2}{4b}$ gives $y=0,y=4a^{2/3}{b}^{1/3}$.

Thus, the two curves intersect at point $P(4a^{1/3}{b}^{2/3},4a^{2/3}{b}^{1/3})$ other than origin $O(0,0)$.
Now, $y^2=4ax$ and $x^2=4by$
Differentiate above equations w.r.t. $x$,
$\Rightarrow 2y\frac{dy}{dx}=4a$ and $2x=4b\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{2a}{y}$ and $\frac{dy}{dx}=\frac{x}{2b}$

$m_1=\frac{dy}{dx}=\frac{2a}{y}=\frac{2a}{4a^{2/3}{b}^{1/3}}=\frac{1}{2}{\left(\frac{a}{b}\right)}^{1/3}$ and
$m_2=\frac{dy}{dx}=\frac{x}{2b}=\frac{4a^{1/3}{b}^{2/3}}{2b}=2{\left(\frac{a}{b}\right)}^{1/3}$

Let $\theta$ be the angle between the tangents to the parabolas $y^2=4ax$ and $x^2=4by$ at P. Then,
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
$=\left|\frac{\frac{1}{2}{\left(\frac{a}{b}\right)}^{1/3}-2{\left(\frac{a}{b}\right)}^{1/3}}{1+\frac{1}{2}{\left(\frac{a}{b}\right)}^{1/3}\times 2{\left(\frac{a}{b}\right)}^{1/3}}\right|$
$=\left|\frac{\frac{-3}{2}{\left(\frac{a}{b}\right)}^{1/3}}{1+{\left(\frac{a}{b}\right)}^{2/3}}\right|$
$\tan \theta =\left|\frac{3a^{1/3}{b}^{1/3}}{2(a^{2/3}+b^{2/3})}\right|$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{3(ab{)}^{1/3}}{2(a^{2/3}+b^{2/3})}\right)$