Find the angle to intersection of the following curves:

Solution:

Given:

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

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

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

Differentiating above w.r.t $x$,

$\Rightarrow 2 y \cdot \frac{d y}{d x}=1$

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

The second curve is $x^{2}=y$

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

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

Substituting (1) in (2), we get

$\Rightarrow x^{2}=y$

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

$\Rightarrow y^{4}-y=0$

$\Rightarrow y\left(y^{3}-1\right)=0$

$\Rightarrow y=0$ or $y=1$

Substituting $y=0 \& y=1$ in $(1)$ in $(2)$,

$x=y^{2}$

when $y=0, x=0$

when $y=1, x=1$

Substituting above values for $m_{1} \& m_{2}$, we get,

when $x=0$

$m_{1}=\frac{d y}{d x}=\frac{1}{2 \times 0}=\infty$

when $x=1$

$m_{1}=\frac{d y}{d x}=\frac{1}{2 \times 1}=\frac{1}{2}$

Values of $m_{1}$ is $\infty \& \frac{1}{2}$

when $y=0$ ,

$\mathrm{m}_{2}=\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}=2 \times 0=0$

when $x=1$

$m_{2}=\frac{d y}{d x}=3 x=2 \times 1=2$

Values of $m_{2}$ is $0 \& 2$

when $m_{1}=\infty \& m_{2}=0$

$\tan \theta=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|$

$\tan \theta=\left|\frac{0-\infty}{1+\infty \times 0}\right|$

$\tan \theta=\infty$

$\theta=\tan ^{-1}(\infty)$

$\therefore \tan ^{-1}(\infty)=\frac{\pi}{2}$

$\theta=\frac{\pi}{2}$

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

$\tan \theta=\left|\frac{2-\frac{1}{2}}{1+\frac{1}{2} \times 2}\right|$

$\tan \theta=\left|\frac{\frac{2}{2}}{2}\right|$

$\tan \theta=\left|\frac{3}{4}\right|$

$\theta=\tan ^{-1}\left(\frac{3}{4}\right)$

$\theta \equiv 36.86$