Hello, poorm87!
Let $\displaystyle C$ be a circle with center $\displaystyle \left(\frac{a}{2},\,0\right)\text{ and radius }\frac{a}{2}$
Let $\displaystyle L$ be the line with equation $\displaystyle x=a$
Find the polar equation of the curve produced in the following manner:
For every angle $\displaystyle \theta,\; \frac{\pi}{2} < \theta < \frac{\pi}{2}$, consider the ray from the origin
. . that makes the angle $\displaystyle \theta$ with the positive xaxis.
The ray intersects $\displaystyle C$ at point $\displaystyle A$, and $\displaystyle L$ at point $\displaystyle B$. Let $\displaystyle O$ be the origin.
Then the point $\displaystyle P$ on the ray is on the curve if: .$\displaystyle OP \,= \,AB$.
Find the equation for the locus of $\displaystyle P.$ Code:
 
 o B
 / 
 / 
 A / 
 * * o 
 * / * 
 * o * 
* / P *
 / 
* / θ *
  o     *     *  
* *a
 
* *
 * * 
 * * 
 * * * L

We have: .$\displaystyle A(a\cos\theta,\,\theta),\;\;B(a\sec\theta,\,\theta ) $
Then: .$\displaystyle AB \;=\;a\sec\theta  a\cos\theta \;=\;a\left(\frac{1}{\cos\theta}  \cos\theta\right) \;=\;a\left(\frac{1  \cos^2\!\theta}{\cos\theta}\right)$
. . . . . $\displaystyle AB \;=\;a\,\frac{\sin^2\!\theta}{\cos\theta} \;=\;a\cdot\sin\theta\cdot\frac{\sin\theta}{\cos\t heta} \;=\;a\sin\theta\tan\theta $
Hence: .$\displaystyle OP \;=\;AB\;=\;a\sin\theta\tan\theta$
Therefore: . $\displaystyle \boxed{r \;=\;a\sin\theta\tan\theta}$