#### Find the angle of intersection of the curves $y = 4-x^2$  and $y = x^2.$

Given: the curves   $y = 4-x^2$  and $y = x^2.$

To find: the interaction angle between two curves

Explanation: acknowledging the first curve

$y = 4-x^2$

when the above curve is differentiated with respect to X

\begin{aligned} &\frac{d y}{d x}=\frac{d\left(4-x^{2}\right)}{d x}\\ &\frac{d y}{d x}=-2 x=m_{1} \ldots\\ &\text { Considering the second curve }\\ &y=x^{2} \end{aligned}

the second curve differentiated with respect to X

$\frac{d y}{d x}=\frac{d\left(x^{2}\right)}{d x}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}=\mathrm{m}_{2} \ldots (ii)$
Given $y=x^{2}$ Substituting the other curve equation with this
$\\x^{2}=4-x^{2} \\ \Rightarrow 2 x^{2}=4 \\ \Rightarrow x^{2}=2 \\ \Rightarrow x=\pm \sqrt{2}$
When $x=\sqrt{2}$, we get $y=(\sqrt{2})^{2} \Rightarrow y=2$
When $x=-\sqrt{2}$ we get $y=(-\sqrt{2})^{2} \Rightarrow y=2$

Hence the intersection points are $(\sqrt{2}, 2) \ and \ (-\sqrt{2}, 2)$ since angle of intersection can be found using the formula
i.e., $\tan \theta=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|$

Substituting the values from equation (i) and equation (ii), we get

$\Rightarrow \tan \theta=\left|\frac{-2 x-2 x}{1+(-2 x)(2 x)}\right|$$\Rightarrow \tan \theta=\left|\frac{-2 x-2 x}{1+(-2 x)(2 x)}\right|$
$\Rightarrow \tan \theta=\left|\frac{-4 x}{1-4 x^{2}}\right|$
For $(\sqrt{2}, 2),$ the equation gets converted into,

$\\ \Rightarrow \tan \theta=\left|\frac{-4(\sqrt{2})}{1-4(\sqrt{2})^{2}}\right| \\\Rightarrow \tan \theta=\left|\frac{-4(\sqrt{2})}{-7}\right| \\\Rightarrow \theta=\tan ^{-1}\left|\frac{4 \sqrt{2}}{7}\right|$
Hence, the angle at which the curve intersects at $y=4-x^{2} and y=x^{2} is \tan ^{-1}\left|\frac{4 \sqrt{2}}{7}\right|$