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hill0093
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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?
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?