`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 rArr |PF_(2)| = 2 and |PF_(1)| = 4`
Also `b^(2) = a^(2) (1 - e^(2)) rArr 4 = a(1 - a^(2))`
`rArr e = `(sqrt(5))/(3) rArr F_(1) F_(2)` = 2sqrt(5)`
Now `PF_(1)^(2) + PF_(2)^(2) = (F_(1) F_(2))^(2) rArr angle (P) = 90^(@)`
So ,area (`Delta PF_(1)F_(2)) = (1)/(2) (4 xx 2) = 4`