# Find tangent lines of a rotated ellipse through certain point

I am trying to obtain the two tangent lines to an ellipse passing through a certain point. I found some similar questions but none that I could find used the ellipse equation $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F$$.

Looking at the below example, I want to obtain the two points E and F where the ellipse and the point D are known. How am I able to calculate those tangent lines (or points) for a rotated ellipse?

• Commented Jul 2, 2019 at 13:48
• Please delete the algebraic geometry tag. This is not algebraic geometry.
– Con
Commented Jul 2, 2019 at 13:51

Translate the plane so that $$D$$ comes to the origin. The conic equation becomes

$$A(x-x_D)^2+B(x-x_D)(y-y_D)+C(y-y_D)^2+D(x-x_D)+E(y-y_D)+F=0$$

and you can compute the new coefficients.

Now assume that the equation of the tangent is

$$y=mx$$ and you get the condition

$$(cm^2+bm+a)x^2+(em+d)x+f=0.$$

We have tangency when this equation has a double root, i.e.

$$(em+d)^2-4f(cm^2+bm+a)=0.$$

This gives us the solutions

$$m=\frac{\pm\sqrt{(e^2-4cf) (4af-d^2)+(de-2 bf)^2}+2bf-de}{e^2-4cf}.$$

• Ended up using this. Thanks!
– Stan
Commented Jul 9, 2019 at 9:20
• @Stan: glad to know. Not my favorite approach though, as it is not isotropic (and fails for vertical tangents).
– user65203
Commented Jul 9, 2019 at 10:27
1. Purely algebraic approach: from equation of ellipse $$f(x,y)=0$$ and equation of line $$(y-y_0)=\alpha(x-x_0)$$, you can get an quadratic equation: $$\frac{y-y_0}{x-x_0}=-\frac{f'_x}{f'_y}$$ which is the equation of ellipse (which is the same as given one, but scaled and shifted). From this, you can find its intersection with the given ellipse.

2. A better approach. You find canonical parameters of ellipse and thus the affine transformation to transform it to unit circle. You apply this transformation to point $$D\to D'$$ and find its tangent points $$E'$$ and $$F'$$ onto unit circle with simple geometry. Inverse transformation will give you points $$E$$ and $$F$$, since affine transformation doesn't change the tangency property.

Joachimsthal's Notations

• $$s_{ij}=Ax_i x_j+B\left( \dfrac{x_i y_j+x_j y_i}{2} \right)+C y_i y_j+D\left( \dfrac{x_i+x_j}{2} \right)+E\left( \dfrac{y_i+y_j}{2} \right)+F$$

• $$s_{i}=Ax_i x+B\left( \dfrac{y_i x+x_i y}{2} \right)+C y_i y+D\left( \dfrac{x+x_i}{2} \right)+E\left( \dfrac{y+y_i}{2} \right)+F$$

• $$s=Ax^2+Bxy+C y^2+Dx+Ey+F$$

Tangent pairs through $$P(x_1,y_1)$$:

$$s_1^2=s_{11} s$$

See the proofs here and another answer of mine here.