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calculus

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use implcit differenciation to find an eqaution of both the tangent line to the ellipse:
2x^2 + 4y^2 = 36
that passes through the points: 14,3

  • calculus -

    First of all, did you notice that the given point lies outside the ellipse ?
    So from that exterior point there will be two different tangents.

    find dy/dx ....

    4x + 8y dy/dx = 0
    dy/dx = -4x/(8y) = -x/(2y)

    let the point of contact be P(x,y)
    slope of tangent = (y-3)/(x-14)

    so (y-3)/(x-14) = -x/(2y)
    which reduces to
    x^2 + 2y^2 = 14x - 6y (#1)
    the original could be reduced to
    x^2 + 2y^2 = 18

    subtract from #1
    14x - 6y = 18
    7x - 3y = 9
    x = (9+3y)/7
    sub back into x^2 + 2y^2 = 18
    I ended up with
    107y^2 + 54y - 801 = 0
    which actually factored to
    (y+3)(107y - 267) = 0
    y = -3 or y = 267/107
    then x = 18/7 or x = 252/107
    ( I cheated and used my fraction key on my calculator)

    not done yet, have to find the equations ...
    1st tangent:
    slope: sub (-3, 18/7) into -x/(2y) to get
    slope = 6/(36/7) = 7/6

    equation is 7x - 6y = c
    sub in (14,3)
    98 - 18 = c
    c = 80

    first tangent : 7x - 6y = 80

    I will let you do the second, good luck!

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