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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
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The correct answer is:
abe
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
$$
\begin{aligned}
&=a b e(\text { maximum value of } \sin \theta) \\
&=a b e(1)=a b e
\end{aligned}
$$
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
$$
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
$$
\begin{aligned}
&=a b e(\text { maximum value of } \sin \theta) \\
&=a b e(1)=a b e
\end{aligned}
$$
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