Question

From a point $P$ perpendiculars $PM$ and $PN$ are drawn upon two fixed lines which are inclined at an angle $\omega$, and which are taken as the axes of coordinates and meet in $O$; find the locus of $P$ if $MN$ be equal to $2c$.

Answer 1

In $△PMN$

${PM}^2+{PN}^2-2PM.PN\cos \angle NPM={MN}^2......\left(i\right)$

From $△PQM,PM=y\sin \omega$

From $△PRN,PN=x\sin \omega$

In quadrilateral $PMQN$

$\angle Q+\angle M+\angle N+\angle NPM={360}^{\circ }\omega +{90}^{\circ }+{90}^{\circ }+\angle NPM={360}^{\circ }\angle NPM={180}^{\circ }-\omega$

Substituting in $\left(i\right)$

$y^2{\sin }^2\omega +x^2{\sin }^2\omega -2\left(y\sin \omega \right)\left(x\sin \omega \right)\cos ({180}^{\circ }-\omega )=4c^2{y}^2{\sin }^2\omega +x^2{\sin }^2\omega +2xy{\sin }^2\omega \cos \omega =4c^2{x}^2+y^2+2xy\cos \omega =4c^2{\csc }^2\omega$

Hence proved.