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Math Help - Normal line to a curve

  1. #1
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    Normal line to a curve

    x^2 + 2xy - 3y^2 = 0

    The line that is normal to this curve at (1,1) intersects the curve at what other point?

    I got the slope to be 1. Can someone tell me how to do this? Please?
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  2. #2
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    Quote Originally Posted by Morgan82 View Post
    x^2 + 2xy - 3y^2 = 0

    The line that is normal to this curve at (1,1) intersects the curve at what other point?

    I got the slope to be 1. Can someone tell me how to do this? Please?
    normal slope is -1

    y-1 = -1(x-1)

    y = -x + 2

    substitute (-x+2) in for y in the original equation ... solve for x, then determine y
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  3. #3
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    If I plug it back in, am I supposed to get (-16+- sqrt(192)) / 8?

    Im still confused.
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  4. #4
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    Quote Originally Posted by Morgan82 View Post
    If I plug it back in, am I supposed to get (-16+- sqrt(192)) / 8?

    Im still confused.
    no ... you should get x = two very nice integer solutions.
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    Which equation do I plug it into?
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  6. #6
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    Quote Originally Posted by skeeter View Post

    substitute (-x+2) in for y in the original equation ...
    ..
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  7. #7
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    oh I got it... a dumb mistake in the solving. Thankyou!
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