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    find the point on the curve

    x^2 + xy + y^2 - 1 = 0

    which is closest to the line x + y = 2

    i differentiated implicitly to find the points where the gradient is equal to that of the line - (1,-1), (-1,1)
    then worked out the perpendicular line from this tangent to the line x + y = 2

    then got the points of intersection as (0,2) and (2,0)

    is this correct?
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    (Original post by jenny0502)
    find the point on the curve

    x^2 + xy + y^2 - 1 = 0

    which is closest to the line x + y = 2

    i differentiated implicitly to find the points where the gradient is equal to that of the line - (1,-1), (-1,1)
    then worked out the perpendicular line from this tangent to the line x + y = 2

    then got the points of intersection as (0,2) and (2,0)

    is this correct?
    No.

    (0 ,2) and (2 ,0) do not lie on that ellipse.

    Find when the gradient of the curve is parallel to the line i.e. when \frac{dy}{dx}=-1 .

    The answer is (\frac{\sqrt{3}}{3},\frac{\sqrt{  3}}{3}) .
 
 
 
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Updated: November 9, 2008

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