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Calculus

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Find equations of both the tangent lines to the ellipse x^2 + 4y^2 = 36 that pass through the point (12, 3).

smaller slope y=
larger slope y=

  • Calculus -

    the the point of contact be P(x,y)
    Make a sketch to see that there would be two points P

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

    slope of tangent = (y-3)/(x-12)
    then
    (y-3)/(x-12) = -x/(4y)
    4y^2 - 12y = -x^2 + 12x
    x^2 + 4y^2 = 12x + 12y
    but x^2 + 4y^2 = 36
    12x + 12y = 36
    x + y = 3
    x = 3-y
    plug back into ellips
    (3-y)^2 + 4y^2 = 36
    9 - 6y + y^2 + 4y^2 = 36
    5y^2 - 6y - 27 = 0
    (y - 3)(5y + 9)
    y = 3 or y = -9/5
    then
    x = 0 or x = 3 - (-9/5) = 24/5
    so P could be (0,3) or P could be (24/5, -9/5)

    case 1: P is (0,3)
    then slope = -x/(4y) = 0/12 = 0
    ahhh, a horizontal line
    y = 3

    case 2 :P is (24/5, -9/5)
    slope = (-24/5) / (4(-9/5)) = 2/3
    so (y + 9/5) = (2/3)(x - 24/5)
    3(y + 9/5) = 2(x - 24/5)
    3y + 27/5 = 2x - 48/5
    3y - 2x = -21/5
    15y - 10x = -21
    10x - 15y = 21

    check my answer for reasonableness by looking at your graph

  • Calculus -

    You can see that tangent from (0,3), the top of the ellipse, to (12,3) is y = 3 by inspection. The other tangent requires some work :)

  • Calculus -

    Hmmm. I get 10x-15y = 75

    http://www.wolframalpha.com/input/?i=plot+x^2%2B4y^2%3D36+and+y%3D3+and+10x-15y%3D75

  • Calculus -

    Steve is right, I made a silly adding vs subtracting error in the 3rd last line

    3y + 27/5 = 2x - 48/5
    3y - 2x = -21/5

    should be

    3y + 27/5 = 2x - 48/5
    3y - 2x = -75/5
    to give
    10x - 15y = 75 , reducing to
    2x - 3y = 15

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