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Suppose \mathrm{F_1} and \mathrm{F_2} are the foci of the ellipse \mathrm{\frac{x^2}{9}+\frac{y^2}{4}=1 . P} is a point on ellipse such that \mathrm{P F_1: P F_2=2: 1}. The area of the triangle \mathrm{P F_1 F_2} is sq. units.

Option: 1

4


Option: 2

6


Option: 3

8


Option: 4

9


Answers (1)

best_answer

\mathrm{e^2=1-\frac{b^2}{a^2}=1-\frac{4}{9}=\frac{5}{9} \Rightarrow e=\frac{\sqrt{5}}{3}}

\mathrm{\begin{aligned} & F_1 F_2=2 \sqrt{5} \\\\ & \text { Also } P F_1+P F_2=6 \\\\ & \text { and } P F_1=2\left(P F_2\right)(\text { Given }) \\\\ & \therefore \quad 3 P F_2=6 \\\\ & \Rightarrow P F_2=2 \\\\ & \text { and } P F_1=4 \\\\ & \text { Since }\left(P F_2\right)^2+\left(P F_1\right)^2=\left(F_1 F_2\right)^2 \\\\ & \Rightarrow \angle P=90^{\circ} \\\\ & \text { Area }=\frac{4 \cdot 2}{2}=4 \text { sq. units } \end{aligned}}

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Ritika Jonwal

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