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Math Help - About tangent to the ellipse

  1. #1
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    [SOLVED] About tangent to the ellipse

    Given the equation of the ellipse x^2+12y^2=48
    Find the equations of the tangents drawn from the point (4,6) and thier points of contact with the ellipse.

    I got the gradient of the tangent which is \frac{\,dy}{\,dx}=-\frac{x}{12y}
    Then I equate it with the gradient of the line
    -\frac{x}{12y}=\frac{y-6}{x-4}

    -x(x-4)=12y(y-6)

    12y^2+x^2=72y+4x

    72y+4x=48 Here is when Im stuck
    Last edited by imforumer; November 12th 2008 at 03:35 AM.
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  2. #2
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    Quote Originally Posted by imforumer View Post
    Given the equation of the ellipse x^2+12y^2=48
    Find the equations of the tangents drawn from the point (4,6) and thier points of contact with the ellipse.

    I got the gradient of the tangent which is \frac{\,dy}{\,dx}=-\frac{x}{12y}
    Then I equate it with the gradient of the line
    -\frac{x}{12y}=\frac{y-6}{x-4}

    -x(x-4)=12y(y-6)

    12y^2+x^2=72y+4x

    72y+4x=48 Here is when Im stuck
    You've done well. Now that can easily be reduced to x= 12- 18y= 6(2- 3y). Put that into the equation for the ellipse and you have a single quadratic equation to solve for y.
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  3. #3
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    but what the straight line 72y+4x=48 stands for?
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