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 $2c$.

Answer 1

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 $OM+ON=2c$

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

$h+k\cos \omega +k+h\cos \omega =2ch+k+\cos \omega (h+k)=2c(h+k)(1+\cos \omega )=2c(h+k)(1+2{\cos }^2\frac{\omega }{2}-1)=2ch+k=\frac{c}{{\cos }^2\frac{\omega }{2}}h+k=c{\sec }^2\frac{\omega }{2}$

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

$x+y=c{\sec }^2\frac{\omega }{2}$

is the required locus of $P$