question_answer

Let P be a variable point on the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] with foci \[{{\operatorname{F}}_{1}}\,\,and\,\,{{F}_{2}}\]. If A is the area of the triangle \[P{{F}_{1}}{{F}_{2}}\], then maximum value of A is

A) ab                    

B)        abe

C) \[\frac{e}{ab}\]             

D)        \[\frac{ab}{e}\]

Correct Answer: B

Solution :

Consider, \[\operatorname{a} > b\]   Area of PF. \[P{{F}_{1}}{{F}_{2}}=\frac{1}{2}({{F}_{1}}\,{{F}_{2}})\times PL\] \[=\,\,\frac{1}{2}2ae\times y=ae.\frac{b}{a}\sqrt{{{a}^{2}}-{{x}^{2}}}\] \[\left[ \begin{align}   & \because \,\,\,\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1 \\  & \Rightarrow \,\,\,y=b\,\sqrt{1-\frac{{{x}^{2}}}{{{a}^{2}}}} \\ \end{align} \right]\] \[A=eb\sqrt{{{a}^{2}}-{{x}^{2}}}\], which is maximum when \[\operatorname{x}= 0\]. Thus the maximum value of A is abe. Similarly, we can solve for \[\operatorname{b} > a.\]