question_answer

If P is a point on \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]with foci S and s?, then the maximum value of \[\Delta SPS'\]is

A)  ab                                         

B) \[ab{{e}^{2}}\]

C)  \[abc\]                                

D)  ab/e

Correct Answer: C

Solution :

Given equation of an ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]                           ??(i) The area of \[\Delta \,SPS'=\frac{1}{2}\left[ \begin{matrix}    ae & 0 & 1  \\    a\cos \theta  & b\sin \theta  & 1  \\    -ae & 0 & 1  \\ \end{matrix} \right]\] Expanding w.r. to \[{{C}_{2}}\] \[=\frac{1}{2}(b\,\sin \theta )\,(ae+ae)\] \[=abe\,\sin \theta \] \[[\because \,-1\le \sin \theta \le 1]\] Here,   \[S'\to (-ae,0)\]                 \[S\to (ae,0)\] Now, maximum value of area = abe (maximum value of \[\sin \theta \]) \[=abe\,(1)=abe\]