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Thread: Finding polar equations

  1. #1
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    Finding polar equations

    I'm having some trouble with this, please help!

    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!!
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  2. #2
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    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: . $\displaystyle \boxed{r \;=\;a\sin\theta\tan\theta}$

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  3. #3
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    Quote Originally Posted by Soroban View Post


    \;=\;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 ??
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