Find the area of the triangle whose vertices are $\left(a \cos \theta , b \sin \theta\right)$, $\left(-a \sin \theta, b \cos \theta\right)$ and $\left(-a \cos \theta, -b \sin \theta\right)$
Let $\;$ $A \left(x_1, y_1\right) = \left(a \cos \theta , b \sin \theta\right)$, $\;$ $B \left(x_2, y_2\right) = \left(-a \sin \theta, b \cos \theta\right)$, $\;$ $C \left(x_3, y_3\right) = \left(-a \cos \theta, -b \sin \theta\right)$
Area of $\triangle ABC = \dfrac{1}{2} \left[x_1 \left(y_2 - y_3\right) + x_2 \left(y_3 - y_1\right) + x_3 \left(y_1 - y_2\right)\right]$
$= \dfrac{1}{2} \left[a \cos \theta \left(b \cos \theta + b \sin \theta\right) - a \sin \theta \left(-b \sin \theta - b \sin \theta\right) - a \cos \theta \left(b \sin \theta - b \cos \theta\right)\right]$
$= \dfrac{1}{2} \left[ab \cos^2 \theta + ab \sin \theta \cos \theta + 2ab \sin^2 \theta -ab \sin \theta \cos \theta + ab \cos^2 \theta\right]$
$= \dfrac{1}{2} \left[2ab \cos^2 \theta + 2 ab \sin^2 \theta\right]$
$= \dfrac{1}{2} \left[2ab \left(\sin^2 \theta + \cos^2 \theta\right)\right]$
$= \dfrac{1}{2} \times 2ab \times 1 = ab$