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    • Thread Starter

    (1+2)(2+y) = x^2 + y^2

    find dy/dx in terms of x and y (Done that bit)

    dy/dx = 2x - 2 - y / x + 1 -2y

    then find the gradient of the above equation at each of the two points where the curve meets the y-axis (Also done this bit)

    dy/dx = 4/3 and -1/3

    then show also that there are two points at which the tangents to this curve are parallel to the y-axis. Not got much of a clue on this bit could someone help out? A gradient parallel to the x axis is 0..but then whats the one for parallel to y? because then i could substitute that value in as dy/dx to calculate a value in terms of x's and y's surely? and they would be the points..?

    When the tangent is parallel to the y-axis, dy/dx = infinity.

    Parallel to x-axis => dy/dx = 0
    Parallel to y-axis => dx/dy = 0
    • Thread Starter

    Thankyou I've got the value for x using that method

    5 +- 2Root13 / 3

    but now struggling for Y. To get x i did dy/dx = 0. therefore x+1-2y = 0
    and x+1 / 2 = y which i then substitued into the first equation. to get a quadratic.

    Now do i just do 2y - 1 = x? because that doesnt seem to work out for me when i substitute it in..please help again Thanks

    Nevermind, simple maths error..duh :\ and only 5 or so weeks to the exam

    Thanks for ya help
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Updated: April 14, 2006
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