Question

If co-ordinates of $P$ and $Q$ are $\left(a\cos \theta ,b\sin \theta \right)$ and $\left(-a\sin \theta ,b\cos \theta \right)$ respectively , then show that $O{P}^2+O{Q}^2=a^2+b^2,$ where $O$ is origin.

Answer 1

$P=(a\cos \theta ,b\sin \theta )Q=(-a\sin \theta ,b\cos \theta )$

$0$ is the origin

$\therefore O{P}^2+O{Q}^2=[\sqrt{(0-a\cos \theta {)}^2+(0-b\sin \theta {)}^2}{]}^2+[\sqrt{(0+a\cos \theta {)}^2+(0-b\sin \theta {)}^2}{]}^2$

$=a^2{\cos }^2\theta +b^2{\sin }^2\theta +a^2{\sin }^2\theta +b^2{\cos }^2\theta$

$=a^2[{\cos }^2\theta +{\sin }^2\theta ]+b^2[{\cos }^2\theta +{\sin }^2\theta ]$

$=a^2+b^2$