Question

Find the locus of the points of intersection of lines x. cos $\propto +y=aandxsin\propto -ycos\propto =bwhere\propto$ variable.

Answer 1

We have,

$x\cos \alpha +y\sin \alpha =a.......\left(1\right)$

$x\cos \alpha -y\sin \alpha =b......\left(2\right)$

On squaring and adding both equation

${(x\cos \alpha +y\sin \alpha )}^2+{(x\sin \alpha -y\cos \alpha )}^2=a^2+b^2$

$\Rightarrow x^2{\cos }^2\alpha +y^2{\sin }^2\alpha +2xy\sin \alpha \cos \alpha +x^2{\sin }^2\alpha +y^2{\cos }^2\alpha -2xy\sin \alpha \cos \alpha =a^2+b^2$

$\Rightarrow x^2{\cos }^2\alpha +x^2{\sin }^2\alpha +y^2{\sin }^2\alpha +y^2{\cos }^2\alpha =a^2+b^2$

$\Rightarrow x^2({\cos }^2\alpha +{\sin }^2\alpha )+y^2({\cos }^2\alpha +{\sin }^2\alpha )=a^2+b^2$

$\Rightarrow x^2+y^2=a^2+b^2$

Hence, this is the answer.