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If $a>0, b>0$ then the maximum area of the parallelogram whose three vertices are $O(0,0), A(a \cos \theta, b \sin \theta)$ and $\mathrm{B}(\mathrm{a} \cos \theta,-\mathrm{b} \sin \theta)$ is
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The correct answer is:
ab when $\theta=\pi / 4$
Area $\mathrm{OABC}=2 \times$ area $\triangle \mathrm{OAB}$
$\begin{array}{l}
=2 \times\left(\frac{1}{2} \times \mathrm{OM} \times \mathrm{AB}\right) \\
=2 \times \frac{1}{2}|\mathrm{a} \cos \theta \times 2 \mathrm{~b} \sin \theta|=2 \times \mathrm{ab}|\sin \theta \cos \theta| \\
=\mathrm{ab}|\sin 2 \theta|
\end{array}$
$\therefore$ Maximum is ab when $\theta=\pi / 4$
$\begin{array}{l}
=2 \times\left(\frac{1}{2} \times \mathrm{OM} \times \mathrm{AB}\right) \\
=2 \times \frac{1}{2}|\mathrm{a} \cos \theta \times 2 \mathrm{~b} \sin \theta|=2 \times \mathrm{ab}|\sin \theta \cos \theta| \\
=\mathrm{ab}|\sin 2 \theta|
\end{array}$
$\therefore$ Maximum is ab when $\theta=\pi / 4$
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