Conic Sections Question 463
Question: Let P be a variable point on the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ with foci $ F_1 $ and $ F_2 $ . If A is the area of the triangle $ PF_1F_2 $ , then maximum value of A is
[IIT 1994; Kerala (Engg.) 2005]
Options:
ab
abe
C) $ \frac{e}{ab} $
D) $ \frac{ab}{e} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ b\sqrt{a^{2}-b^{2}} $ if $ a>b;a\sqrt{b^{2}-a^{2}} $ if $ P $ Area of $ PF_1F_2=\frac{1}{2}(F_1F_2)\times PL $
$ =\frac{1}{2}(2ac)\times y=ae\frac{b}{a}\sqrt{a^{2}-x^{2}} $
$ A=eb\sqrt{a^{2}-x^{2}} $ , which is maximum when $ x=0 $ . Thus the maximum value of A is abe.
BETA
