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Thread: About tangent to the ellipse

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

    Given the equation of the ellipse $\displaystyle x^2+12y^2=48$
    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 $\displaystyle \frac{\,dy}{\,dx}=-\frac{x}{12y}$
    Then I equate it with the gradient of the line
    $\displaystyle -\frac{x}{12y}=\frac{y-6}{x-4}$

    $\displaystyle -x(x-4)=12y(y-6)$

    $\displaystyle 12y^2+x^2=72y+4x$

    $\displaystyle 72y+4x=48$ Here is when Im stuck
    Last edited by imforumer; Nov 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 $\displaystyle x^2+12y^2=48$
    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 $\displaystyle \frac{\,dy}{\,dx}=-\frac{x}{12y}$
    Then I equate it with the gradient of the line
    $\displaystyle -\frac{x}{12y}=\frac{y-6}{x-4}$

    $\displaystyle -x(x-4)=12y(y-6)$

    $\displaystyle 12y^2+x^2=72y+4x$

    $\displaystyle 72y+4x=48$ 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 $\displaystyle 72y+4x=48$ stands for?
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