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I apologize for the horrible title. I came across this in an exam question: You're given $C_1$ as $$r = 1 + cos 2\theta $$ For $\frac{\pi}{2} \leq \theta \leq -\frac{\pi}{2}$. The only symmetry $C_1$ resembles is reflection about the polar axis.
It intersects with $C_2$, defined as $$r = 1 + sin \theta $$ , at the point $A$, which works out to be $(\frac{3}{2}, \frac{\pi}{6})$.

Then, a line is introduced, which is parallel to the polar axis, goes through $A$, and meets $C_1$ at a second point, $B$. Now you're supposed to show that the length of $OB$† equals some value, from which I only remember that it includes $\sqrt{13}$.
How is this solved? I.e. what is the exact length of $OB$ in surd form?

Intuitively, I tried to work out the angle by equating the distance from $A$ to the polar axis ($\frac{3}{4}$) with $r \sin \theta$, which is the vertical component of a polar point: $$(1 + \cos 2\theta)\sin \theta = \frac{3}{4}$$ Which can be transformed to $${\cos^2 \theta} ~\sin \theta = \frac{3}{8}$$ , which doesn't really yield anything simple - a depressed cubic in terms of $sin \theta$.

$O$ is the pole

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    $\begingroup$ What is "the initial line"? There doesn't seem to be any other line defined ... $\endgroup$ – hmakholm left over Monica May 13 '15 at 22:39
  • $\begingroup$ @HenningMakholm It seems I found the correct term: polar axis. $\endgroup$ – Columbo May 13 '15 at 22:44
  • $\begingroup$ $\sin\theta=z$, $\frac{(1-z^2)z-\frac{3}{8}}{z-\frac{1}{2}}$$=-\frac{1}{4} (4 z^2+2 z-3)$ -- quadratic :) , $\sin\theta= \frac{1}{4} (\sqrt{13}-1)$ $\endgroup$ – Alexey Burdin May 13 '15 at 22:59
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    $\begingroup$ @AlexeyBurdin Oh well. I totally missed that I could've done polynomial division while looking for an elegant trigonometrical solution... that's embarassing. $\endgroup$ – Columbo May 13 '15 at 23:02
  • $\begingroup$ Incidentally, if you plot the curve you will find there are three possible choices for point $B$, one of which has $|OB|=|OA|=\frac32$. $\endgroup$ – hmakholm left over Monica May 14 '15 at 15:18
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I would start by rewriting the equation for $C_1$ in mixed Cartesian and polar coordinates:

$$ r = 1+\cos2\theta = 2\cos^2\theta = 2- 2\sin^2\theta = 2-2\frac{y^2}{r^2}$$ whence $$ r^3 - 2r^2 + 2y^2 = 0 $$

Your line $AB$ is defined as having $y=\frac32\sin\frac\pi6=\frac34$, which you can insert in this equation and get an equation for $r$. You also already know one root, namely $r=\frac32$, which you can divide out and use polynomial division to find the other roots (one of which will be negative).

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