Let $F_{1},F_{2}$ are foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1;P$ is a point on th

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Let $F_{1},F_{2}$ are foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1;P$ is a point on the ellipse such that $\frac{PF_{1}}{PF_{2}} = 2$. Then the area of $\bigtriangleup PF_{1}F_{2}$ is:

Ans. (d)

Accepted Answer

Sol. Given ellipse: x²/9 + y²/4 = 1Here, a = 3, b = 2, so c = √(a² − b²) = √5Hence, the foci are F₁(−√5, 0) and F₂(√5, 0).Given: PF₁/PF₂ = 2Also, for any point P on the ellipse,PF₁ + PF₂ = 2a = 6Therefore, PF₁ = 4, PF₂ = 2Now, in ΔPF₁F₂: F₁F₂ = 2c = 2√5Using Heron's formula,s = (4 + 2 + 2√5)/2 = 3 + √5Area = √[s(s − 4)(s − 2)(s − 2√5)]= √[(3 + √5)(√5 − 1)(1 + √5)(3 − √5)]= √[(9 − 5)(5 − 1)]= √16= 4

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