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 $PM+PN$ be equal to $2c$.

Answer 1

Let the point $P$ be $(h,k)$

Draw $P{L}^{\prime }$ parallel $MO$ and $PL$ parallel $NO$

In $△PLM\sin \omega =\frac{PM}{PL}$

$\sin \omega =\frac{PM}{k}\Rightarrow PM=k\sin \omega ......\left(i\right)$

In $△P{L}^{\prime }N\sin \omega =\frac{PN}{P{L}^{\prime }}$

$\sin \omega =\frac{PN}{h}\Rightarrow PN=h\sin \omega ......\left(ii\right)$

Given $PM+PN=2c$

using $\left(i\right)$ and $\left(ii\right)$

$h\sin \omega +k\sin \omega =2ch+k=\frac{2c}{\sin \omega }h+k=2c\csc \omega$

Replacing $h$ by $x$ and $y$ by $k$

$x+y=2c\csc \omega$