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Math Help - Finding the intersection between the normal line and an implicit irrational function

  1. #1
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    Finding the intersection between the normal line and an implicit irrational function

    PROBLEM:

    Intersecting normal:
    The line that is normal to the curve x^2+2xy-3y^2=0 at (1,1) intersects the curve at what other point?

    ATTEMPT:

    My biggest problem lies in finding the intersection between a single-valued function and a multi-valued function. Finding the intersection between a single valued function and the same is easy enough, so I figured if I could express
    x^2+2xy-3y^2=0 as a series of single valued functions, then I should be able to find where the normal intersects. At least in theory. With some investigation, I realized I didn't know how to properly express the curve, x^2+2xy-3y^2=0, as such. I thought that using the quadratic formula might be appropriate for this operation since the curve is a two-valued function specifically, which resulted in the expression: y=\frac{x}{3}\pm\sqrt{16}x. But when I tried finding the pair, it resulted in (\frac{7}{53},\frac{343}{159}) and (\frac{7}{43},\frac{329}{129}) which doesn't agree with what I expected from the graph.

    I've determined that the normal curve is the function, y=-\frac{4}{3}x+\frac{7}{3}.

    I appreciate any thoughts you have on this.
    Thanks for reading.

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  2. #2
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    Re: Finding the intersection between the normal line and an implicit irrational funct

    \frac{d}{dx}(x^2 + 2xy - 3y^2 = 0)

    \frac{dy}{dx} = \frac{x+y}{3y-x}

    at (1,1) , \frac{dy}{dx} = 1

    normal line thru (1,1) is ...

    y - 1 = -1(x-1)

    y = 2-x

    substitute (2-x) for y in the original curve equation ...
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  3. #3
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    Re: Finding the intersection between the normal line and an implicit irrational funct

    Algebra error:
    y={2x\pm \sqrt{4x^2+12x^2}\over 6}={x\over 3}\pm {2x\over 3}
    As pointed out above, the normal line has equation y=2-x. Easy algebra shows the other point of intersection is (3,-1).
    Finding the intersection between the normal line and an implicit irrational function-calculus101.png
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  4. #4
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    Re: Finding the intersection between the normal line and an implicit irrational funct

    You can also do it like this
    x2+2xy-3y2=0
    x2+2xy+y2-y2-3y2=0
    (x+y)2-4y2=0
    (x+y)2=(2y)2
    The normal line to x=y and passing through (1,1) has the equation y=2-x
    y=2-x intersects x=-3y at (3,-1)
    Thus the required point is (3,-1)
    x+y=2y
    x=y, x=-3y
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