Question

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

Answer 1

We have $y=4-x^2...\left(i\right)$
and $y=x^2....\left(ii\right)$
$\Rightarrow \frac{dy}{dx}=-2x$
and $\frac{dy}{dx}=2x$
$\Rightarrow m_1=-2x$
and $m_2=2x$
From eqs (i) and (ii) $x^2=4-x^2$
$\Rightarrow 2x^2=4$
$\Rightarrow x^2=2$
$\Rightarrow x=\pm \sqrt{2}$
$\therefore y=x^2=(\pm \sqrt{2}{)}^2=2$

So the points of intersection are $(\sqrt{2},2)$ and $(-\sqrt{2},2)$

For point $(+\sqrt{2},2)$

$m_1=-2x=-2\sqrt{2}=-2\sqrt{2}$

and $m_2=2x=2\sqrt{2}$

and for point $(\sqrt{2},2)$

$tan\theta =|\frac{m_1-m_2}{1+m_1{m}_2}|=|\frac{-2\sqrt{2}-2\sqrt{2}}{1-2\sqrt{2}2\sqrt{2}}|=|\frac{-4\sqrt{2}}{-7}|$

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