If tangents are drawn from a point P(2,0) to the curve √1+y2=x3 which meet the curve at A and B, then
√1+y2=x3
Differentiating w.r.t. x,
2y2√1+y2(dydx)=13
⇒dydx=√1+y23y
dydx∣∣∣(h,k)=√1+k23k
∴k−0h−2=√1+k23k
⇒3k2h−2=√1+k2
⇒3(h29−1)h−2=h3
⇒h2−9h−2=h
⇒h2−9=h2−2h
⇒h=92
So, k=±√52
Let AP make angle θ with x−axis in anticlockwise direction.
Then, acute angle between the tangents is 2θ.
Slope of AP is tanθ=1√5
tan2θ=2tanθ1−tan2θ=√52
⇒2θ=tan−1(√52)
AB=√5
Area of △PAB is 12×√5×(92−2)=5√54
AP≠AB
Hence, △PAB is not equilateral.