A variable line $L$ through $P(3,4)$ meets lines $x-y+6=0$ and $x-y+10=0$ at $A$ and $B$. A point $Q$ is on $L$ such that $PQ^2=PA.PB$ . Find locus of $Q$.
Attempt:
If variable line $L$ makes angle $\theta$ with the axis, equation of Line $ L $ is
$\frac{x-3}{cos \theta}$=$\frac{y-4}{sin \theta}$=$r1,r2$
Where $r1$ and $r2$ are distances from lines $ x-y+6=0$ and $x-y+10=0$.
I have got distances as function of $ \theta$ . After this I am stuck. How will I eliminate this parameter? Please help. If I am proceeding in wrong direction then please suggest an appoach.
Best Answer
(it is important do draw a picture to see in particular that the straight lines are parallel).
Let us consider $P$ as the new origin. Doing that, we must operate a change of coordinates ; the formulas are (old coordinates expressed as functions of the new ones) :
$$\begin{cases}x=X+3\\ y=Y+4\end{cases},$$
Therefore, the equations of lines $(L_1)$ and $(L_2)$ become :
$$\begin{cases}(L_1): & X-Y+5=0 \\ (L_2): & X-Y+9=0\end{cases},$$
Taking $X=r\cos \theta, Y=r\sin \theta$, cartesian equation of $(L_1)$ above becomes polar equation $r\cos \theta-r\sin \theta+5=0$, otherwise said,
$$r=\dfrac{5}{-\cos \theta+\sin \theta}=\dfrac{5}{\sqrt{2}\cos (\theta - 3 \pi/4)}$$
For $(L_2)$ :
$$r=\dfrac{9}{\sqrt{2}\cos (\theta - 3 \pi/4)}$$
Otherwise said, the polar equations of these lines with respect to point $P$ (taken as origin) are of the form :
$$p=PA=\dfrac{p_1}{\cos(\theta-\theta_0)} \ \ \ \text{and} \ \ \ p=PB=\dfrac{p_2}{\cos(\theta-\theta_0)}\tag{1}$$
for $\theta_0=3 \pi/4, \ p_1=\dfrac{5}{\sqrt{2}}, \ p_2=\dfrac{9}{\sqrt{2}}$.
(formulas in (1) are the most classical for the polar equations of straight lines with $r$ taking its minimal value (corresponding to shortest distance points) when the denominator is maximal, i.e. when $\theta=\theta_0$.
Remark : the fact that it is the same $\theta_0$ for both lines means that they are parallel .
Therefore relationship $PQ=\sqrt{PA.PB}$, using (1), gives the polar equation of the locus of point $Q$ (still with $P$ considered a the origin):
One can work "backwards" to find a cartesian equation for this locus but it is not at all necessary.