If $P$ is a point on $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with focii $S$ and $S^{\prime}$, then the maximum value of $\triangle S P S^{\prime}$ is

• A

  $a b$
• B

  $a b e^{2}$
• Correct answer
  C

  abe
• D

  $a b / e$

Solution

Given equation of an ellipse $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $$ The area of $\Delta S P S^{\prime}=\frac{1}{2}\left[\begin{array}{ccc}a e & 0 & 1 \\ a \cos \theta & b \sin \theta & 1 \\ -a e & 0 & 1\end{array}\right]$ Expanding w.r, to $C_{2}$ $$ \begin{aligned} &=\frac{1}{2}(b \sin \theta)(a e+a e) \\ &=a b e \sin \theta[\because-1 \leq \sin \theta \leq 1] \end{aligned} $$

Here, $S^{\prime} \rightarrow(-a e, 0)$ $$ S \rightarrow(a e, 0) $$ Now, maximum value of area $$ \begin{aligned} &=a b e(\text { maximum value of } \sin \theta) \\ &=a b e(1)=a b e \end{aligned} $$

Hence, the correct answer is option: 'C'.