P is a variable point on ellipse 9x2+16y2=144 with foci S and S′. If K is the area of triangle SS′P, then the maximum value of K2 is
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Solution
Equation of the ellipse is 9x2+16y2=144⇒x216+y29=1 ∴a=4,b=3 and e=√1−916=√74 Distance between foci SS′=2ae=2√7 Any point on the ellipse is P(4cosθ,3sinθ) ∴ Area of △SS′P=12×SS′×3sinθ ⇒K=3√7sinθ ⇒K2=63sin2θ ∴K2max=63