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March 25, 2017

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Show that the tangents to the curve X=ln((y^2)-6y+11) ,at the points where X=ln6,meet at the point ((ln6)-(4/3),3)

  • A level Mathematics - ,

    first of all we need dy/dx for the relation, that would be

    1 = (2y(dy/dx) - 6dy/dx)/(y^2 - 6y + 11)

    simplifying and solving for dy/dx gave me
    dy/dx = (y^2 - 6y + 11)/(2y-6)

    then let's sub in x=ln6 into the original
    ln6 = ln(y^2 - 6y + 11)
    6 = y^2 - 6y + 11
    solving as a quadratic, it factored, I got
    y = 1 or y = 5

    so at the point (ln6,1), dy/dx = -3/2
    and at point (ln6,5), dy/dx = +3/2

    first tangent:
    y-1 = (-3/2)(x-ln6) which reduced to
    3x + 2y = 2 + 3ln6 (#1)

    second tangent:
    y-5 = (3/2)(x-ln6) gave me
    3x - 2y = 3ln6 - 10 (#2)

    adding #1 and #2 gave me

    x = ln6 - 4/3

    and carefully subbing that back into #1 gave me y = 3

    YES!!!!!

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