Straight line involving polar form

Question

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.

Answer 1

(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):

$$p=PQ=\dfrac{\sqrt{p_1p_2}}{\cos(\theta-\theta_0)}\tag{2}$$

which is the equation of a straight line parallel to the initial lines (see remark above) and situated in between them (but not at equal distance from them).

One can work "backwards" to find a cartesian equation for this locus but it is not at all necessary.

Answer 2

Guide:

• Write an equation of this line $L$: $$y-4=k(x-3)$$ and calulate it intersection with other two line, you get $A$ and $B$ (as variable points depending on $k$).

• Let $C$ be a mipoint of $AB$ and write an equation of (variable) circle with diameter $AB$. Note that $C$ is on fixed line in the midlle of a given lines.

• Write an equation of (variable) circle with diameter $PC$ and calculate it interesection points with first circle. You get say points $M,N$.

• Calculate angle $\angle CPM =\phi$ and see it is constant, so $C$ maps to $M$ under spiral similarity, so $M$ describe some line $\ell$

• Finally, rotate $M$ for angle $\phi$ to $Q$, so $Q$ describe also some line which is picture of $\ell$ under this rotation (actualy it is parallel to $x-y+10=0$).

Answer 3

Before proving the particular case you asked, it may be instructive to state a general proposition or theorem:

Between two parallel lines $ (y=a,y=b) $ if a pencil of rays is drawn through origin then the line of geometric means of extreme intercepts as shown is the parallel straight line $ y= \sqrt{ab}.$

The proof of the above considers relevant similar triangles from picture below and the locus straight line is marked red.

In a longer derivation involving polar coordinates for our situation the locus is a straight line

$$ y= m x +a\quad y= m x + b \tag1$$

Translate to $ (h,k) = (3,4) $ ; plug into the first line $$ (y-k) = m (x-h) +a $$ and slope of line through origin is $\dfrac{y}{x}=p$. The $x$ coordinate $$ x= \frac{a+k-m h}{p-m}=\frac{Q}{p-m} \tag2 $$ similarly $$ y=\frac{Qp}{p-m} \tag3 $$ square and add, take square root $$ r =Q \frac{\sqrt{1+p^2}}{p-m} \tag4 $$ $p=\tan \theta$ $$ r= \frac{Q \sec \theta}{\tan \theta -m}=\frac{Q }{\sin \theta -m \cos \theta} \tag5$$ cross multiply $$ y-m x=Q = a+k-m h \tag6$$ similarly $$ y-mx=R = b+k-m h $$ Now convenient to convert to polar form $$ r_A= Q \frac {\sec \theta}{p-m};\,r_B= R \frac {\sec \theta}{p-m};\tag7 $$ so as to obtain its geometric mean $$ r= \frac{\sqrt {QR}\,\sec \theta} {p-m} \tag8$$ Calculate Cartesian coordinates $$ (x,y)= \frac{\sqrt{QR} (1,p)}{p-m} \tag9 $$ simplifying with $p = \tan \theta = y/x$ $$ y-mx= (mh-k)+ \sqrt{(a+k-mh)(b+k-mh)} $$ $$ y = (m x - C) + \sqrt{(a+C)(b+C)} ,\quad C= ( k-m h)\tag{10} $$

where

$$ (h,k,m)=(3,4,1) $$