Angle of intersection of curves $y=4-x^2$ and $y=x^2$ is

detailed solution

Correct option is C

First of all we note that the two curvesy=4−x2 andx=y meet where y=4−y⇒y=2 and hencex2=2⇒x=±2x . So, the two curves meet at the points(2,2) ,(−2,2) . We shall find the angle of intersection at(2,2) . For x2=y, dydx=2x andFory=4−x2, dydx=−2x .Hence, the slopes of the tangents to the two curves at (2,2) are 22 and −22 respectively.∴   Required angle of intersection=tan−1(|22+221+22(−22)|)=tan−1(427) .