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Thread: Polar Curve Geometry

  1. #1
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    Polar Curve Geometry

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


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


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

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

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

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

    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 $\displaystyle (r,\theta)$ format.

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

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

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

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

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

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

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

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