1. ## Finding polar equations

Let C be a circle with center (a/2,0) and radius a/2. Let l be the line with equation x=a. Find the polar equation of the curve produced in the following manner: for every angle theta, -pi/2 < theta < pi/2, consider the ray from the origin that makes the angle theta with the positive x-axis. This ray intersects C at a point A and l at point B and let O be the origin. Then the point P on the ray is on the curve if the segment OP = the segment AB.

I can see what the curve looks like (sort of an exponential line) but I cannot figure out how to find the equation for it. thanks for the help!!

2. 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 x-axis.

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: . $\displaystyle \boxed{r \;=\;a\sin\theta\tan\theta}$

3. Originally Posted by Soroban

\;=\;a\left(\frac{1 - \cos^2\!\theta}{\cos\theta}\right)[/tex]

. . . . . [tex]AB \;=\;a\,\frac{\sin^2\!\theta}{\cos\theta}

I don't follow this step, is 1-cos^2theta = sin^2theta ??