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 $OM-ON$ be equal to $2d$.

Answer 1

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

Draw $P{L}^{\prime }$ parallel $OM$ and $PL$ parallel $ON$

In $△PLM\cos \omega =\frac{LM}{PL}$

$\Rightarrow LM=PL\cos \omega =k\cos \omega OM=OL+LMOM=h+k\cos \omega ........\left(i\right)$

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

$\Rightarrow L^{\prime }N=P{L}^{\prime }\cos \omega =h\cos \omega ON=O{L}^{\prime }+L^{\prime }NON=k+h\cos \omega ......\left(ii\right)$

Given $OM-ON=2d$

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

$h+k\cos \omega -k-h\cos \omega =2dh-k-\cos \omega (h-k)=2d(h-k)(1-\cos \omega )=2d(h-k)(1-(1-2{\sin }^2\frac{\omega }{2}))=2d(h-k)2{\sin }^2\frac{\omega }{2}=2dh-k=d{\csc }^2\frac{\omega }{2}$

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

$x-y=d{\csc }^2\frac{\omega }{2}$