Question

Find the relation obtained by eliminating $\theta$ from the equations $x=r$ $\cos \theta +s\sin \theta$ and $y=r$$\sin \theta -s\cos \theta$

Answer 1

Given, $x=r$ $\cos \theta +s\sin \theta$
$\Rightarrow x^2={(r\cos \theta +s\sin \theta )}^2=r^2{\cos }^2\theta +2rs\cos \theta .\sin \theta +s^2{\sin }^2\theta$
Also

$y=r\sin \theta -s\cos \theta \Rightarrow y^2=r^2{\sin }^2\theta +s^2{\cos }^2\theta -2rs\sin \theta \cos \theta$
$\Rightarrow x^2+y^2=r^2({\cos }^2\theta +{\sin }^2\theta )+s^2({\sin }^2\theta +{\cos }^2\theta )$
$=r^2\left(1\right)+s^2\left(1\right)[\because {\sin }^2\theta +{\cos }^2\theta =1]=r^2+s^2$
Hence, the required relation is $x^2+y^2=r^2+s^2$