question_answer

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 \[P{{F}_{1}}{{F}_{2}}\], then maximum value of A is [IIT 1994; Kerala (Engg.) 2005]

A)            ab   

B)            abe

C)            \[\frac{e}{ab}\]                         

D)            \[\frac{ab}{e}\]

Correct Answer: B

Solution :

           \[b\sqrt{{{a}^{2}}-{{b}^{2}}}\]if \[a>b;a\sqrt{{{b}^{2}}-{{a}^{2}}}\]if \[P\] Area of \[P{{F}_{1}}{{F}_{2}}=\frac{1}{2}({{F}_{1}}{{F}_{2}})\times PL\]                     \[=\frac{1}{2}(2ac)\times y=ae.\frac{b}{a}\sqrt{{{a}^{2}}-{{x}^{2}}}\]                   \[A=eb\sqrt{{{a}^{2}}-{{x}^{2}}}\], which is maximum when\[x=0\].                    Thus the maximum value of A is abe.