Help in understanding a proof about polar coordinates and locus.

Question

A variable straight line is drawn through a given point $O$ to cut two fixed lines at $R$ and $S$; we take $P$ on such line in order that $${2 \over OP} = {1 \over OR} + {1\over OS}$$ holds. We have to show that the locus of $P$ is a third fixed straight line.

---

Let $O$ be the origin of the rectangular axes such that $OX$ is x axis and $OY$ is the y axis,

Then the equation to fixed lines are $Ax + By + C = 0$ and $A_0x+B_0y +C_0 = 0$

In polar coordinates the eqautions are then

$$ {1\over r} = -{A\cos\theta + B\sin\theta\over C} \text{ and } {1\over r} = -{A_0\cos\theta + B_0\sin\theta\over C_0} \tag{1}$$

If $\angle XOR = \theta$ then,

$${1\over OR} = -{A\cos\theta + B\sin\theta\over C} \tag{2}$$ $${1\over OS} = -{A_0\cos\theta + B_0\sin\theta\over C_0} \tag{3}$$

$$\therefore {2\over OP} = -{A\cos\theta + B\sin\theta\over C} -{A_0\cos\theta + B_0\sin\theta\over C_0}$$

Therefore the locus of $P$ on transfroming back to cartesian coordinates is

$$2 = -x\left({A\over C} + {A_0 \over C_0}\right) -y\left({B\over C} + {B_0 \over C_0}\right)$$

Hence proved.

---

I did not understand $(1),(2)$ and $(3)$,

• Although I know the general equation for a line in polar coordinates is $r\cos(\theta -\alpha) = p$ where, $\alpha$ and $p$ are the angle between the line and the $\perp$ from the origin on the line, I did not get from this how we get the form given in the proof.

• In $(2), (3)$ I did not understand why $\angle XOR = \theta \implies 1/OR , 1/OS = 1/r$

---

Answer 1

1) This form of the polar equation of a line is obtained by substituting $x=r\cos\theta$ and $y=r\sin\theta$ in the cartesian equation $Ax+By+C=0$.

2) If $\angle XOR=\theta$, then plugging this angle into the polar equation of the first line will give $r=OR$, whence this equality.

3) As $O$, $R$ and $S$ are on the same line, we have either $\angle XOS=\angle XOR=\theta$ or $\angle XOS=\pi+\angle XOR=\pi+\theta$. In your proof only the first case is considered, but the second case can be handled in a similar way, because $\cos(\pi+\theta)=-\cos\theta$ and $\sin(\pi+\theta)=-\sin\theta$. But the two lines you get are different in the two cases, and in general you get both of them: a segment of the first one and a segment of the second one.

In addition, notice that on the variable line there are two points satisfying the relation for $P$. That gives another couple of segments. These four segments form a parallelogram.

Answer 2

In the above configuration, if we consider a unit circle centered at $O$ and the circle inversion $x\mapsto x^{-1}$, both $s^{-1}$ and $r^{-1}$ are circles through $O$ and $(s\cap r)^{-1}$. $P^{-1}$ is just the midpoint of the segment delimited by $OP\cap s^{-1}$ and $OP\cap r^{-1}$, hence $P^{-1}$ lies on a circle through $O$ and $(s\cap r)^{-1}$, too. By inverting $P^{-1}$ we get that $P$ lies on a line through $s\cap r$.

In a simpler way, we may notice that the given constraint fixes a cross ratio among the four points $S,P,R,O$.