KCET · Maths · Ellipse
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 \(a b e^{2}\)
- C abe
- D \(a b / e\)
Answer & Solution
Correct Answer
(C) abe
Step-by-step Solution
Detailed explanation
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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