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  4. If a>0, b>0 then the maximum area of the parallelogram whose three vertices are O(0,0) ⋅ A(a cos θ, b sin θ) and B(a cos θ,-b sin θ) is

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Q. If $a>0, b>0$ then the maximum area of the parallelogram whose three vertices are $O(0,0) \cdot A(a \cos \theta, b \sin \theta)$ and $B(a \cos \theta,-b \sin \theta)$ is

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Solution:

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Area $O A B C=2 \times$ area $\triangle O A B$
$=2 \times\left(\frac{1}{2} \times O M \times A B\right)$
$=2 \times \frac{1}{2}|a \cos \theta \times 2 b \sin \theta|$
$=2 \times a b|\sin \theta \cos \theta|$
$=a b|\sin 2 \theta|$
$\therefore $ Maximum is ab when $\theta=\pi / 4$