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Math Help - Focal length

  1. #1
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    Focal length

    hi,
    I have to find the focal length of x^2-6x+8y-7=0
    I had already found the turning point using calculus. (3,2)

    I've never done it from this starting point but I rearranged it to get this

    \begin{align*}x^2-6x+9&=-8x-16\\(x-3)^2&=-8(y-2)\\(x-3)^2&=4\times-2(y-2)\end{align*}

    which reaffirms that the vertex is (3,2) and gives the focal length of a=-2

    Is this the BEST way of doing it, especially considering that I had already used calculus to find the vertex?
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  2. #2
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    Re: Focal length

    Quote Originally Posted by Melody2 View Post
    hi,
    I have to find the focal length of x^2-6x+8y-7=0
    I had already found the turning point using calculus. (3,2)

    I've never done it from this starting point but I rearranged it to get this

    \begin{align*}x^2-6x+9&=-8x-16\\(x-3)^2&=-8(y-2)\\(x-3)^2&=4\times-2(y-2)\end{align*}

    which reaffirms that the vertex is (3,2) and gives the focal length of a=-2

    Is this the BEST way of doing it, especially considering that I had already used calculus to find the vertex?
    There is a fast way. If you can get your equation into the form y=a x^2 + b x + c then the focal length is f=\frac{1}{4a}

    in this case you have to rearrange things a bit to obtain

    y=-\frac{1}{8}\left(x^2-6x-7\right) and thus read off a as -\frac{1}{8} so f=-2
    Thanks from Melody2
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  3. #3
    Senior Member sakonpure6's Avatar
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    Re: Focal length

    nvm
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