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 $MN$ pass through the fixed point $(a,b)$.

Answer 1

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

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

In $△PLM,\cos \omega =\frac{LM}{PL}$

$OM=OL+LMOM=h+k\cos \omega$

So, the coordinates of $M$ are $(h+k\cos \omega ,0)$

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

$\Rightarrow L^{\prime }N=P{L}^{\prime }\cos \omega =h\cos \omega ON=O{L}^{\prime }+L^{\prime }NON=k+h\cos \omega$

So the coordinates of $N$ are $(0,k+h\cos \omega )$

Equation of $MN$ is

$y-0=\frac{k+h\cos \omega -0}{0-(h+k\cos \omega )}(x-h-k\cos \omega )y=-\frac{k+h\cos \omega }{h+k\cos \omega }(x-h-k\cos \omega )-yh-yk\cos \omega =kx-kh-k^2\cos \omega +hx\cos \omega -h^2\cos \omega -hk{\cos }^2\omega h^2\cos \omega +k^2\cos \omega +hk(1+{\cos }^2\omega )=h(y+x\cos \omega )+k(x+y\cos \omega )$

It passes through $(a,b)$

$\Rightarrow h^2\cos \omega +k^2\cos \omega +hk(1+{\cos }^2\omega )=h(b+a\cos \omega )+k(a+b\cos \omega )$

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

$x^2\cos \omega +y^2\cos \omega +xy(1+{\cos }^2\omega )=x(b+a\cos \omega )+y(a+b\cos \omega )$

is the required locus of $P$