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Least Circumradius of Triangle PS1S2 on an Ellipse | JEE

JEE Maths question with a full step-by-step solution.

Question
Let PP be a point on the ellipse x2a2+y2b2=1 (a>b)\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\ (a>b) in the first quadrant, whose foci are S1S_1 and S2S_2. Then the least possible value of the circumradius of △PS1S2\triangle PS_1S_2 is:
Aaeaecorrect
Bbebe
Caeb\dfrac{ae}{b}
Dae2b\dfrac{ae^2}{b}
Solution
Step 1: In △PS1S2\triangle PS_1S_2, the side S1S2=2aeS_1S_2=2ae is opposite the angle ∠P=∠S1PS2\angle P=\angle S_1PS_2. By the sine rule for the circumradius,
R=S1S22sin⁡(∠P)=2ae2sin⁡(∠P)=aesin⁡(∠P).R=\frac{S_1S_2}{2\sin(\angle P)}=\frac{2ae}{2\sin(\angle P)}=\frac{ae}{\sin(\angle P)}.
 Step 2: Since sin⁡(∠P)≤1\sin(\angle P)\le1,
R=aesin⁡(∠P)≥ae,R=\frac{ae}{\sin(\angle P)}\ge ae,
 with equality when ∠P=π2\angle P=\dfrac{\pi}{2}. ∴\therefore the least possible value of the circumradius is aeae. Correct answer: (1)
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