Find the angle to intersection of the following curves:

Solution:

Given:

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

$\& x^{2}+y^{2}=20 \ldots(2)$

First curve $y=x^{2}$

$\Rightarrow \mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x} \ldots$ (3)

Second curve is $x^{2}+y^{2}=20$

Differentiating above w.r.t $x$,

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

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

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

Substituting $(1)$ in $(2)$, we get

$\Rightarrow y+y^{2}=20$

$\Rightarrow y^{2}+y-20=0$

We will use factorization method to solve the above Quadratic equation

$\Rightarrow y^{2}+5 y-4 y-20=0$

$\Rightarrow y(y+5)-4(y+5)=0$

$\Rightarrow(y+5)(y-4)=0$

$\Rightarrow y=-5 \& y=4$

Substituting $y=-5 \& y=4$ in $(1)$ in $(2)$,

$y=x^{2}$

when $y=-5$

$\Rightarrow-5=x^{2}$

$\Rightarrow x=\sqrt{-5}$

when $y=4$

$\Rightarrow 4=x^{2}$

$\Rightarrow x=\pm 2$

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

when $x=2$

$m_{1}=\frac{d y}{d x}=2 \times 2=4$

when $x=1$

$\mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=2 \times-2=-4$

Values of $m_{1}$ is $4 \&-4$

when $y=4 \& x=2$

$\mathrm{m}_{2}=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{x}}{\mathrm{y}}=\frac{-2}{4}=\frac{-1}{2}$

when $y=4 \& x=-2$

$m_{2}=\frac{d y}{d x}=\frac{-x}{y}=\frac{2}{4}=\frac{1}{2}$

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

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

$\tan \theta=\left|\frac{\frac{-1}{2}-4}{1+2 \times 4}\right|$

$\tan \theta=\left|\frac{\frac{-9}{2}}{1-2}\right|$

$\tan \theta=\left|\frac{9}{2}\right|$

$\theta=\tan ^{-1}\left(\frac{9}{2}\right)$

$\theta \cong 77.47$