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Math Help - Polar Curve Geometry

  1. #1
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    Polar Curve Geometry

    Q: Sketch the curve with equation r=a(1+\cos\theta) for 0\le\theta\le\pi where a>0. Sketch also the line with equation r=2a\sec\theta for -\frac{\pi}{2}<\theta<\frac{\pi}{2} on the same diagram.
    The half-line with equation \theta=\alpha, 0<\alpha<\frac{\pi}{2}, meets the curve at A and the line with equation r=2a\sec\theta at B. If O is the pole, find the value of \cos\alpha for which OB=2OA.


    I drew the sketch but can't do second part. Could someone help? Thanks in advance.
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  2. #2
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    Quote Originally Posted by Air View Post
    Q: Sketch the curve with equation r=a(1+\cos\theta) for 0\le\theta\le\pi where a>0. Sketch also the line with equation r=2a\sec\theta for -\frac{\pi}{2}<\theta<\frac{\pi}{2} on the same diagram.
    The half-line with equation \theta=\alpha, 0<\alpha<\frac{\pi}{2}, meets the curve at A and the line with equation r=2a\sec\theta at B. If O is the pole, find the value of \cos\alpha for which OB=2OA.


    I drew the sketch but can't do second part. Could someone help? Thanks in advance.
    We will see all points in (r,\theta) format.

    Clearly A = (a(1+\cos \alpha),\alpha) and B = (2a \sec \alpha, \alpha).

    OA = a(1+\cos \alpha) and OB = 2a \sec \alpha.

    So OB = 2OA \Rightarrow 2a \sec \alpha = 2a(1+\cos \alpha) \Rightarrow \cos^2 \alpha + \cos \alpha - 1 = 0

    Solving the quadratic equation, discarding the invalid root, we can get our answer...



    P.S: Well I am not sure of it but it looks correct, if you have any doubts, just ask ok?It will help me too
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  3. #3
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    Quote Originally Posted by Isomorphism View Post
    We will see all points in (r,\theta) format.

    Clearly A = (a(1+\cos \alpha),\alpha) and B = (2a \sec \alpha, \alpha).

    OA = a(1+\cos \alpha) and OB = 2a \sec \alpha.

    So OB = 2OA \Rightarrow 2a \sec \alpha = 2a(1+\cos \alpha) \Rightarrow \cos^2 \alpha + \cos \alpha - 1 = 0

    Solving the quadratic equation, discarding the invalid root, we can get our answer...



    P.S: Well I am not sure of it but it looks correct, if you have any doubts, just ask ok?It will help me too

    How did you get the co-ordinates?
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by Air View Post
    Q: Sketch the curve with equation r=a(1+\cos\theta) for 0\le\theta\le\pi where a>0. Sketch also the line with equation r=2a\sec\theta for -\frac{\pi}{2}<\theta<\frac{\pi}{2} on the same diagram.
    The half-line with equation \theta=\alpha, 0<\alpha<\frac{\pi}{2}, meets the curve at A and the line with equation r=2a\sec\theta at B. If O is the pole, find the value of \cos\alpha for which OB=2OA.
    I drew the sketch but can't do second part. Could someone help? Thanks in advance.
    Quote Originally Posted by Air View Post
    How did you get the co-ordinates?
    The question says : "The half-line with equation \theta=\alpha, 0<\alpha<\frac{\pi}{2}, meets the curve at A"

    So the point A where both the curves meet must satisfy both \theta=\alpha, 0<\alpha<\frac{\pi}{2} and r=a(1+\cos\theta) for 0\le\theta\le\pi where a>0.

    So substitute \theta=\alpha in r=a(1+\cos\theta), to get r. We already know that \theta=\alpha. Thus we can get (r,\theta).

    Do a similar thing for B by finding intersection with r=2a\sec\theta.

    Now OA is the distance from the pole to the point A. That is nothing but the radial distance...
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