Find the angle of intersection of the curves y=4−x2and y=x2
We have y=4−x2
and y=x2⇒dydx=−2xand dydx=2x⇒m1=−2xand m2=2xfrom Eqs.(i) and (ii)x2=4−x2⇒2x2=4⇒x2=2⇒x=±√2∴y=x2=(±√2)2=2
So, the points of intersection are (√2,2) and (−√2,2).
For point (+√2,2),m1=−2x=−2.√2=−2√2
and m2=2x=2√2.
For point (√2,2),tanθ=∣∣m1−m21+m1m2∣∣=∣∣∣−2√2−2√21−2√2.2√2∣∣∣=∣∣∣−4√2−7∣∣∣θ=tan−1(4√27)