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Math Help - Polar Equation Help

  1. #1
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    Unhappy Polar Equation Help

    Find the polar equation that has the same graph as the given Cartesian.

    x^2-12y-36=0

    I can't figure out how to do this problem.
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  2. #2
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    Quote Originally Posted by dachosin1 View Post
    Find the polar equation that has the same graph as the given Cartesian.

    x^2-12y-36=0

    I can't figure out how to do this problem.
    Start by substituting x = r \cos \theta and y = r \sin \theta.

    Show your all working and say where you get stuck.
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  3. #3
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    x^2-12y-36=0

    (rcosx)^2-12rsinx-36=0

    (r^2)(cos(x)^2)-12rsinx-36=0

    (r^2)(cos(x)^2)-12rsinx=36

    rcos(x)^2-12sinx=36/r

    rcos(x)^2=(36/r) + 12sinx

    rcos(x)^2=(36/sqrt(x^2+y^2))+12sinx

    r=36/[sqrt(x^2+y^2)(cos(x)^2)]+12sinx/cos(x)^2

    is this the last step?
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  4. #4
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    Quote Originally Posted by dachosin1 View Post
    x^2-12y-36=0

    (rcosx)^2-12rsinx-36=0

    (r^2)(cos(x)^2)-12rsinx-36=0

    (r^2)(cos(x)^2)-12rsinx=36

    rcos(x)^2-12sinx=36/r

    rcos(x)^2=(36/r) + 12sinx

    rcos(x)^2=(36/sqrt(x^2+y^2))+12sinx

    r=36/[sqrt(x^2+y^2)(cos(x)^2)]+12sinx/cos(x)^2

    is this the last step?
    You are meant to get r in terms of \theta. That means no y's or x's.

    You have (\cos^2 \theta) r^2 - (12 \sin \theta) r - 36 = 0. This is a quadratic equation in r. Solve for r using the quadratic formula.
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  5. #5
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    Perhaps we don't need to solve it by quadratic formula , juz take square root

    x^2 - 12y - 36 = 0 <=>
    r^2(cos^2 a) - 12r(sina) - 36 = 0 <=>
    r^2[ 1 -(sin^2a)] - 12r(sina) - 36 = 0 <=>
    r^2 - { [rsin(a)]^2 + 2*6[rsin(a)] + 6^2 } = 0 <=>
    r^2 = [rsin(a) + 6]^2 <=>
    r = rsin(a) + 6 OR r = -rsin(a) + 6
    r = 6/[1 - sin(a)] OR r = 6/[ 1+ sin(a)]
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