If a0, b0, then the maximum area of the par | vedclass

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(A) Let the vertices of the parallelogram be $O(0,0)$, $A(a \cos 	heta, b \sin 	heta)$, $B(a \cos 	heta, -b \sin 	heta)$, and $C(x, y)$.Since $OAB

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$ab$ when $	heta=\pi / 4$

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(A) Let the vertices of the parallelogram be $O(0,0)$, $A(a \cos 	heta, b \sin 	heta)$, $B(a \cos 	heta, -b \sin 	heta)$, and $C(x, y)$.Since $OABC$ is a parallelogram, the diagonals $OB$ and $AC$ bisect each other at the same midpoint.Alternatively, the area of a parallelogram with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and $(x_4, y_4)$ can be calculated using the cross product of vectors.The area of $	riangle OAB = \frac{1}{2} |x_A y_B - x_B y_A| = \frac{1}{2} |(a \cos 	heta)(-b \sin 	heta) - (a \cos 	heta)(b \sin 	heta)| = \frac{1}{2} |-2ab \sin 	heta \cos 	heta| = ab |\sin 	heta \cos 	heta| = \frac{ab}{2} |\sin 2	heta|$.The area of the parallelogram $OABC = 2 	imes 	ext{Area}(	riangle OAB) = 2 	imes \frac{ab}{2} |\sin 2	heta| = ab |\sin 2	heta|$.The maximum value of $|\sin 2	heta|$ is $1$, which occurs when $2	heta = \pi / 2$, i.e.,$	heta = \pi / 4$.Thus, the maximum area is $ab$ when $	heta = \pi / 4$.

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 $B(a \cos \theta, -b \sin \theta)$ is

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