Find the angle to intersection of the following curves :

Solution:

Given:

Curves $x^{2}=27 y \ldots(1)$

$\& y^{2}=8 x \ldots(2)$

Solving $(1) \&(2)$, we get,

From $y^{2}=8 x$, we get,

$\Rightarrow x=\frac{y^{2}}{8}$

Substituting $x=\frac{y^{2}}{8}$ on $x^{2}=27 y$,

$\Rightarrow\left(\frac{y^{2}}{8}\right)^{2}=27 y$

$\Rightarrow\left(\frac{y^{4}}{64}\right)=27 y$

$\Rightarrow y^{4}=1728 y$

$\Rightarrow y\left(y^{3}-1728\right)=0$

$\Rightarrow y=0$ or $\left(y^{3}-1728\right)=0$

$\Rightarrow y=0$ or $y=\sqrt[3]{1728}$

$\therefore \sqrt[3]{1728}=12$

$\Rightarrow y=0$ or $y=12$

Substituting $y=0$ or $y=12$ on $x=\frac{y^{2}}{8}$

when $y=0$

$\Rightarrow x=\frac{0^{2}}{8}$

$\Rightarrow x=0$

$\therefore$ The point of intersection of two curves $(0,0) \&(18,12)$

First curve is $x^{2}=27 y$

Differentiating above w.r.t $\mathrm{x}$,

$\Rightarrow 2 x=27 \cdot \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{2 x}{27}$

$\Rightarrow m_{1}=\frac{2 x}{27} \ldots(3)$

Second curve is $y^{2}=8 x$

$\Rightarrow 2 y \cdot \frac{d y}{d x}=8$

$\Rightarrow y \cdot \frac{d y}{d x}=4$

$\Rightarrow m_{2}=\frac{4}{y} \ldots(4)$

Substituting $(18,12)$ for $m_{1} \& m_{2}$, we get,

$m_{1}=\frac{2 x}{27}$

$\Rightarrow \frac{2 \times 18}{27}=\frac{36}{27}$

$m_{1}=\frac{4}{3} \ldots(5)$

$m_{2}=\frac{4}{y}$

$\Rightarrow \frac{4}{y}=\frac{4}{12}$

$m_{2}=\frac{1}{3} \ldots(6)$

when $m_{1}=\frac{4}{3} \& m_{2}=\frac{1}{3}$

$\Rightarrow \tan \theta=\left|\frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{3} \times \frac{1}{3}}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{3}{3}}{1+\frac{4}{9}}\right|$

$\Rightarrow \tan \theta=\left|\frac{1}{\frac{12}{9}}\right|$

$\Rightarrow \tan \theta=\left|\frac{9}{13}\right|$

$\Rightarrow \theta=\tan ^{-1}\left(\frac{9}{13}\right)$

$\Rightarrow \theta \equiv 34.69$