Question

Find the angle of intersection of the following curve:
$y=4-x^2$ and $y=x^2$

Answer 1

$y=4-x^2$, $y=x^2$

Solving the two curves, we get

$4-x^2=x^2$

$2x^2=4$

$x^2=2$

$x=\pm \sqrt{2}$

$x=\sqrt{2}$ $y=2$

$x=-\sqrt{2}$ $y=2$

for $\left(m_1\right),\frac{dy}{dx}=-2x$

for $\left(m_2\right),\frac{dy}{dx}=2x$

at $(\sqrt{2},2)$ $m_1=-2\sqrt{2}{m}_2=2\sqrt{2}$

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

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

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

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