Find the angle to intersection of the following curves :

Solution:

Given:

Curves $y=4-x^{2} \ldots(1)$

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

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

$\Rightarrow y=4-x^{2}$

$\Rightarrow x^{2}=4-x^{2}$

$\Rightarrow 2 x^{2}=4$

$\Rightarrow x^{2}=2$

$\Rightarrow x=\pm \sqrt{2}$

Substituting $\sqrt{2}$ in $y=x^{2}$, we get

$y=(\sqrt{2})^{2}$

$y=2$

The point of intersection of two curves are $(\sqrt{2}, 2) \&(-\sqrt{2},-2)$

First curve $y=4-x^{2}$

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

$\Rightarrow \frac{d y}{d x}=0-2 x$

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

Second curve $y=x^{2}$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}$

$m_{2}=2 x \ldots(4)$

At $(\sqrt{2}, 2)$, we have,

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

$\Rightarrow-2 \times \sqrt{2}$

$\Rightarrow m_{1}=-2 \sqrt{2}$

At $(-\sqrt{2}, 2)$, we have,

$m_{2}=\frac{d y}{d x}=-2 x$

$(-1) \times-\sqrt{2} \times 2=2 \sqrt{2}$

When $m_{1}=-2 \sqrt{2} \& m_{2}=2 \sqrt{2}$

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

$\tan \theta=\left|\frac{-4 \sqrt{2}}{1-8}\right|$

$\tan \theta=\left|\frac{-4 \sqrt{2}}{-7}\right|$

$\tan \theta=\frac{4 \sqrt{2}}{7}$

$\theta=\tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)$

$\theta \cong 38.94$