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Math Help - help solving for y

  1. #1
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    help solving for y

    (y/7)^2 + ((x-1)/3)^2 = 1

    How do I go about doing this?
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  2. #2
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    Quote Originally Posted by clambottomjewels View Post
    (y/7)^2 + ((x-1)/3)^2 = 1

    How do I go about doing this?
    This is the equation of an ellipse:

    \left(\dfrac y7\right)^2+\left(\dfrac{x-3}3\right)^2=1~\implies~\dfrac{(x-3)^2}{9}+\dfrac{y^2}{49}=1

    So what exactly do you want to do with it?

    EDIT: OK, finally I read the headline

    \dfrac{(x-3)^2}{9}+\dfrac{y^2}{49}=1 ~\implies~y^2=49-\dfrac{49}9 \cdot (x-3)^2

    Can you take it from here?
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  3. #3
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    Quote Originally Posted by clambottomjewels View Post
    (y/7)^2 + ((x-1)/3)^2 = 1

    How do I go about doing this?
    • Take \left(\frac{1}{3}(x-1)\right)^2 from both sides
    • Take the square root of both sides (and don't forget the \pm
    • Multiply both sides by 7


    You should get an answer of y = \pm7\sqrt{1-\left(\frac{1}{3}(x-1)\right)^2}. This is equivalent to

    \pm \frac{7}{3} \sqrt{-x^2+2x+8}
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