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Math Help - Line intersecting implicit function

  1. #1
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    Line intersecting implicit function

    Hi, I have to find the co-ordinates where the line y=-x+1 intersects f(x,y) = x^2 + y^2 -xy - y -1

    I can do this with explicit functions by making them equal, but it doesn't work with this.

    The actual question asks for the co-ordinates where the tangent to f(x,y), is at 45 deg to the horizontal. So I found the gradient function which = 1 (that got me the linear equation), now i'm stuck.

    Not sure if this should be in calc. section, so sorry if I got it wrong.
    Thanks, Rich.
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  2. #2
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    Hello hciR
    Quote Originally Posted by hciR View Post
    Hi, I have to find the co-ordinates where the line y=-x+1 intersects f(x,y) = x^2 + y^2 -xy - y -1

    I can do this with explicit functions by making them equal, but it doesn't work with this.

    The actual question asks for the co-ordinates where the tangent to f(x,y), is at 45 deg to the horizontal. So I found the gradient function which = 1 (that got me the linear equation), now i'm stuck.

    Not sure if this should be in calc. section, so sorry if I got it wrong.
    Thanks, Rich.
    I assume that the curve is f(x,y)=0, in which case I agree that y = 1-x when \frac{dy}{dx}=1.

    Just substitute y = 1-x into f(x,y)=0 to eliminate y, and solve the resulting quadratic in x. I make the answer
    x=-\tfrac13, \;1
    Then find the corresponding values of y.

    Grandad
    Last edited by Grandad; January 18th 2010 at 10:04 AM. Reason: Corrected calculation
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  3. #3
    MHF Contributor ebaines's Avatar
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    The line y = -x + 1 lies in the x-y plane, or where f(x,y) = 0.

    So sub that in:
    0 = x^2 + y^2 - xy -y -1,
    0 = x^2 + (-x+1)^2 -x(-x+1) -(-x+1) -1

    Now solve for x.

    EDIT - I got:

    <br />
 x = \frac {2 \pm \sqrt {16} } 6 = 1, -\frac 1 3<br />
    Last edited by ebaines; January 18th 2010 at 09:53 AM. Reason: added info
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  4. #4
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    Oh ok, i was going about it the wrong way thanks.

    As an aside:- I managed to get those co-ordinates for x earlier in the question, where it asked for the stationary points. Is it just the nature of the shape that the x co-ordinates are the same, when the gradient is 0 and 45 deg?

    Also, when working out the S.P I got four possible Y values, and only knew which ones to use by plotting the graph, is there another mathematical way of doing it?
    Thanks.
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