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Two tangent points of two lines from (0,0) to an ellipse

  1. Nov 12, 2011 #1
    How do I find the two tangent points of two lines from (0,0) to an ellipse?

    We have 2 equations, a general ellipse and it differentiated:
    1: A*x*x+B*x*y+C*y*y+D*x+E*y+F=0 is an ellipse if B*B-4*A*C<0.
    Differentiating, 2: 2*A*x+B*x*dy/dx+B*y+2*C*y*dy/dx+D+E*dy/dx=0.

    If F<0, ellipse not on (0,0), line from (0,0) tangent to ellipse is:
    3: y=dy/dx*x where dy/dx is that of the ellipse. So 3:dy/dx=y/x.

    Rearranging equation 2:
    2: dy/dx=-(2*A*x+B*y+D)/(2*C*y+B*x+E).

    Combining equation 2 with equation 3:
    4: 2*(A*x*x+B*x*y+C*y*y)+D*x+E*y=0.

    But we haven't satisfied the ellipse equation 1 yet so
    Solving equations 1 and 4 together,
    (D*x+E*y)/2+F=0, but that's not a tangent-point solution.

    How do I get the two solutions of equations 1 and 4?
    Did I make a mistake, or am I doing it wrong?
  2. jcsd
  3. Nov 12, 2011 #2


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    Science Advisor

    There is nothing wrong and you have done extremely well! Now, solve that equation for, say, y and put it into your original equation for the ellipse. That will give you a quadratic equation for x, giving the x coordinate of the two points of tangency.
  4. Nov 12, 2011 #3
    Solving equations 1 and 4 together,
    5: (D*x+E*y)/2+F=0, a line not a tangent-point solution.
    Rearranging by writing y as a function of x,
    5: y=-(x*D/E+2*F/E) and y*y=((D*D/(E*E))*x*x+2*D*F/(E*E)*x+4*F*F/(E*E)

    Substituting into the ellipse equation 1,
    6: A*x*x-B*x*(x*D/E+2*F/E)+C*(((D*D/(E*E))*x*x+2*D*F/(E*E)*x+4*F*F/(E*E))+D*x-E*(x*D/E+2*F/E)+F=0
    Collecting terms for a quadratic to solve using x=(-b+-sqrt(b*b-4*a*c))/(2*a),
    6: (A-B*D/E+C*D*D/(E*E))x*x+2*(C*D*F/(E*E)-B*F/E)*x+(4*F*F/E-F)=0

    I Used the quadratic formula and got two points that I will
    check when I plot and give thus verified code then.

    I also need ellipse tangency points of the two tangent circles about (0,0),
    but will also have to work on that later, and that is more difficult.
    Anyone have the solutions.
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