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## Main Question or Discussion Point

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?