PQ is a chord of the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$.If O is the center of the ellipse and eccentric angle of the points P and Q differ by $\frac{\mathrm{\pi }}{2}$, then the area of triangle OPQ is

$$\begin{gathered} Given ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b) \\ where centre \left(C\right)=(0,0) \\ Let P=(a\cos \theta ,b\sin \theta ) and Q=(-a\sin \theta ,b\cos \theta ) be two points on ellipse \\ Now area of triangle OPQ=\frac{1}{2}\left|a{b}^{ }{\cos }^2\theta +ab{\sin }^2\theta \right| \\ =\frac{ab}{2} \end{gathered}$$