Question

Find the angle of intersection of the curves $y=4-x^{2}andy=x^2$

Answer 1

We have $y=4-x^2$
and $y=x^2\Rightarrow \frac{dy}{dx}=-2xand\frac{dy}{dx}=2x\Rightarrow m_1=-2xand{m}_2=2xfromEqs.\left(i\right)and\left(ii\right)x^2=4-x^2\Rightarrow 2x^2=4\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2}\therefore y=x^2=(\pm \sqrt{2}{)}^2=2$
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 $m_2=2x=2\sqrt{2}$.
For point $(\sqrt{2},2),tan\theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{-2\sqrt{2}-2\sqrt{2}}{1-2\sqrt{2}.2\sqrt{2}}\right|=\left|\frac{-4\sqrt{2}}{-7}\right|\theta =ta{n}^{-1}\left(\frac{4\sqrt{2}}{7}\right)$