Question

A variable line $L$ passing through the point $B(2,5)$ intersects the lines $2x^2-5xy+2y^2=0$ at $P$ and $Q$. Find the locus of the point $R$ on $L$ such that distance $BP,BR$ and $BQ$ are in harmonic progression.

Answer 1

$L$ is $\frac{x-2}{\cos \theta }=\frac{y-5}{\sin \theta }=\begin{array}{c}{r}_1\\ P\end{array}\begin{array}{c}r\\ R\end{array}\begin{array}{c}{r}_2\\ Q\end{array}$
where $r_1,r,r_2$ are in $H.P$.
or $\frac{2}{r}=\frac{1}{r_1}+\frac{1}{r_2}$
${2x}^2-5xy+{2y}^2=0$
or $(2x-y)(x-2y)=0$
$P$ lies on $2x-y=0$,$Q$ on $x-2y=0$
$2\left(r_1\cos \theta +2\right)-(r_1\sin \theta +5)=0$
$r_1(2\cos \theta -\sin \theta )=1$
$\therefore \frac{1}{r_1}=2\cos \theta -\sin \theta$
Similarly $\frac{1}{r_1}=\frac{\cos \theta -2\sin \theta }{8}$
$\therefore \frac{1}{r_1}+\frac{1}{r_2}=\frac{17\cos \theta -10\sin \theta }{8}$
But $\frac{1}{r_1}+\frac{1}{r_2}=\frac{2}{r}=\frac{(17\cos \theta -10\sin \theta )}{8}$
$\therefore 16=(17r\cos \theta -10r\sin \theta )$
If $(x,y)$ be the point $R$, then
$x=r\cos \theta +2$, or $r\cos \theta =x-2$
$y=r\sin \theta +5$
$\therefore r\sin \theta =y-5$
Eliminating $\theta$ between $\left(1\right)$ and $\left(2\right)$, we get
$16=\left[17\right(x-2)-10(y-5\left)\right]=(17x-10y+16)$
or $17x-10y=0$ is the required locus.