The angle of intersection between the curves $y=4-x^2$ and $y=x^2 $ is :

$tan^{-1}\frac{4\sqrt{2}}{7}$

The correct answer is option (4) → $\tan^{-1}\frac{4\sqrt{2}}{7}$

$y=4-x^2$  ...(1)

$y=x^2$  ...(2)

finding intersection point

$4-x^2=x^2⇒x^2=2⇒x=±\sqrt{2}$

$y=2$

from (1) $\frac{dy}{dx}=m=-2x$

from (2) $\frac{dy}{dx}=2x=n$

angle $\tan θ=\left|\frac{m-n}{1+mn}\right|$ at $x=\sqrt{2}$

$θ=\tan^{-1}\left|\frac{2\sqrt{2}+2\sqrt{2}}{1+(-8)}\right|$

$=\tan^{-1}\frac{4\sqrt{2}}{7}$

even with $x = -\sqrt{2}$

$θ=\tan^{-1}=\frac{4\sqrt{2}}{7}$