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 \[\operatorname{b} > a.\]You need to login to perform this action.
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