`F_(1) and F_(2)` are the two foci of the ellipse `(x^(2))/(9) + (y^(2))/(4) = 1.` Let P be a point on the ellipse such that `|PF_(1) | = 2|PF_(2)|`, where `F_(1) and F_(2)` are the two foci of the ellipse . The area of `triangle PF_(1)F_(2)` is :

Options

A. 3
B. 4
C. `sqrt(5)`
D. `(sqrt(13))/(2)`

Answer

B

Explanation

Given ` |PF_1|=2|PF_(2)|`
Also, ` |PF_1 | + |PF_2 | = 6 implies |PF_2| =2 and |PF_1|=4`
Also ` b^(2) = a^(2) (1-e^(2)) implies 4 = a (1-a^(2))`
`implies e=(sqrt(5))/(3) implies F_1F_2 = 2sqrt(5)`

Now `PF_1^(2) + PF_2^(2) = (F_1 F_2)^2 implies angleP= 90^(@)`
So , area `(DeltaPF_1 F_2) = (1)/(2) ( 4xx 2 ) =4 `