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Math Help - tangent of an ellipse

  1. #1
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    tangent of an ellipse

    x^2 + 2x + 4y^2 - 16y + 13v = 0 represents an ellipse

    the line y=mx meets the ellipse

    show that 3m^2 - 4m - 3 => 0.

    What value(s) of m make the line a tangent to the ellipse?
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  2. #2
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    Quote Originally Posted by deej813 View Post
    x^2 + 2x + 4y^2 - 16y + 13v = 0 represents an ellipse

    the line y=mx meets the ellipse

    show that 3m^2 - 4m - 3 => 0.

    What value(s) of m make the line a tangent to the ellipse?
    1. I assume that you mean:

    x^2 + 2x + 4y^2 - 16y + 13 = 0

    2. Calculate the points of intersection between the line and the ellipse substituting y = mx:

    x^2 + 2x + 4m^2 x^2 - 16 m x + 13 = 0 and solve for x. (Use the formula to solve a quadratice equation)

    x = - \dfrac{2\cdot \sqrt{3m^2 - 4m - 3} - 8m + 1}{4m^2 + 1} \vee x =  \dfrac{2\cdot \sqrt{3m^2 - 4m - 3} + 8m - 1}{4m^2 + 1}

    3. You only get one point of intersection = tangent point if the radicand equals zero.

    4. To calculate the values of m, solve for m:

    3m^2 - 4m - 3 = 0 (Use the formula to solve a quadratice equation)
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  3. #3
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    ok sorry still confused
    what are the steps for how you get the equations in 2. the x=-... and x=...
    why are there two??
    how do you solve it for m?
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  4. #4
    ynj
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    Quote Originally Posted by deej813 View Post
    ok sorry still confused
    what are the steps for how you get the equations in 2. the x=-... and x=...
    why are there two??
    how do you solve it for m?
     x^2 + 2x + 4m^2 x^2 - 16 m x + 13 = 0\Leftrightarrow (1+4m^2)x^2+(2-16m)x+13.Then solve it as a quadratic equation.
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  5. #5
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    um ok i tried using the quadratic formula and got to

    [-2(1-8m)+-sqrt(16(3m^2-4m-3))] / 52(1-8m)

    is that right?
    what do i do from here??
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  6. #6
    ynj
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    Quote Originally Posted by deej813 View Post
    um ok i tried using the quadratic formula and got to

    [-2(1-8m)+-sqrt(16(3m^2-4m-3))] / 52(1-8m)

    is that right?
    what do i do from here??
    No, it should be the same answer as earboths'. Then you can prove the first part since a negative number have no square.
    If y=mx is a tangent, it means it has only one intersection with the ellipse. So sqrt(...) must be zero!
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