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Thread: Identifying graph of a polar equation.

  1. #1
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    Identifying graph of a polar equation.

    Having a little trouble with where to go with this problem.

    Identify the graph of the polar equation r = 3/(4-3cosθ)

    3/4/(1-(3/4)cos θ,
    e = -3/4
    ep = 3/4
    3p = 3/4
    p = 1/4

    Did i get these right? and if so how do i go from here to identify the graph?
    I think θ= p/2 at r = 3/4/(1-3/4) = 3 so the graph runs through (p/2,3) right? What from here?
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  2. #2
    Super Member malaygoel's Avatar
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    What are e and p here??

    To identify graph, try to convert in x-y cordinates.

    Spoiler:

    $\displaystyle r=\frac{3}{4-3cos\theta}$

    $\displaystyle 4r-3rcos\theta=3$

    $\displaystyle 4r=3+3x$

    $\displaystyle 16r^2=9+18x+9x^2$

    $\displaystyle 16x^2+16y^2=9+18x+9x^2$

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  3. #3
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    r = 3 / 4 - 3 cos or ep/1 - e cos q
    e would be 3 i think,(or 3/4 after you divide the denominator by 4) and Im not sure if got p right, but if the top is ep, correct me if im wrong but wouldn't it be 3p = 3/4? so p = 1/4?
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  4. #4
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    e is the eccentricity and |p| is the distance between the focus and the directrix.

    Quote Originally Posted by Wesker View Post
    Having a little trouble with where to go with this problem.

    Identify the graph of the polar equation r = 3/(4-3cosθ)
    The form we are interested in is
    $\displaystyle r = \frac{ep}{1 - e\cos \theta}$

    $\displaystyle \begin{aligned}
    r &= \frac{3}{4 - 3\cos \theta} \\
    r &= \frac{0.75}{1 - 0.75\cos \theta}
    \end{aligned}$

    The denominator is $\displaystyle 1 - 0.75\cos \theta$, which means that it has a vertical directrix to the left of the pole.

    e = 3/4 < 1, so the conic is an ellipse.

    The numerator is ep = 3/4, so p = 1 (not 1/4).

    I think θ= p/2 at r = 3/4/(1-3/4) = 3 so the graph runs through (p/2,3) right? What from here?
    No, at $\displaystyle \theta = \pi/2$, $\displaystyle \cos \pi/2 = 0$, so

    $\displaystyle r = \frac{3}{4 - 3\cos (\pi/2)} = \frac{3}{4}$

    From here, you might as well find r at the other quadrantal angles.
    When $\displaystyle \theta = 0$,
    $\displaystyle r = \frac{3}{4 - 3\cos 0} = ????$

    When $\displaystyle \theta = \pi$,
    $\displaystyle r = \frac{3}{4 - 3\cos \pi} = ????$

    When $\displaystyle \theta = 3\pi/2$,
    $\displaystyle r = \frac{3}{4 - 3\cos (3\pi/2)} = ????$


    01
    Last edited by yeongil; Jul 7th 2009 at 10:38 PM. Reason: Typos!
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  5. #5
    Super Member malaygoel's Avatar
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    Quote Originally Posted by yeongil View Post
    When $\displaystyle r = 0$,
    $\displaystyle r = \frac{3}{4 - 3\cos 0} = ????$

    When $\displaystyle r = \pi$,
    $\displaystyle r = \frac{3}{4 - 3\cos \pi} = ????$

    When $\displaystyle r = 3\pi/2$,
    $\displaystyle r = \frac{3}{4 - 3\cos (3\pi/2)} = ????$


    01
    there is typo in the above
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  6. #6
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    Fixed. Thanks.


    01
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