Area of △PQR
$\begin{aligned}
& =\frac{1}{2}(2 a)(a \sin \theta+b) \
& \therefore \text { maximum area }=a(a+b) \
& \quad=4(4+2 \sqrt{3})=8(2+\sqrt{3})
\end{aligned}$
Let C be the circle of minimum area enclosing the ellipse E:a2x2+b2y2=1 with eccentricity 21 and foci (±2,0). Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 29 is parallel to the major axis of E and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is :
Held on 3 Apr 2025 · Verified 6 Jul 2026.
6(3+2)
8(3+2)
62+3
82+3
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