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I've been beating my head against a wall for several days trying to figure this out. It doesn't seem to be a documented problem on the internet anywhere.

I have a horizontal line segment of known length (R). This line is tangent to an ellipse and one endpoint contacts it at the topmost point of the ellipse. At the end of line segment R is another line segment of unknown length. This line segment is at at a known angle off vertical (Θ). This line segment is tangent to the ellipse as well. I also know the "aspect ratio" of the ellipse (B / A).

A is the major dimension and B is the minor dimension. A is parallel to R (i.e. the ellipse is wider than it is tall). A and B are half the overall width and height of the ellipse, respectively.

What I need to calculate are the dimensions of the ellipse (lengths A and B) as well as the location of the contact point on the ellipse of the angled line segment (lengths H and S). Is there a formula for this?

Here is a diagram of what I am talking about

The blue lines in the diagram represent the connection lines from the tangent point to the focii.

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Without loss of generality, place the “top” of the ellipse at the origin. Letting $\rho=b/a$, its equation is $$\rho^2x^2+(y+b)^2=b^2.\tag 1$$ If the line $\lambda x+\mu y+\tau=0$ is tangent to this ellipse, the coefficients of that equation must satisfy the dual conic equation $${b^2\over\rho^2}\lambda^2+2b\mu\tau-\tau^2=0.\tag2$$

As drawn, the other end point of the horizontal segment is at $(-R,0)$, so an equation of the other tangent line is $$x\cos\theta-y\sin\theta+R\cos\theta=0.\tag3$$ Substituting into (2), we get after a bit of rearrangement $$\left(\left(\frac{b^2}{\rho^2}-R^2\right)\cos\theta-2Rb\sin\theta\right)\cos\theta = 0.\tag4$$ This is a straightforward quadratic equation in $b$. Once you know $b$, the point of tangency of the line (3) can be found in a variety of ways. Using pole-polar relationships is easiest since we’ve already worked out the dual conic: the pole of a tangent line to a conic is the point of tangency. The pole of the line $\lambda x+\mu y+\tau=0$ can be found by substitution into the dual conic. For the dual conic (2), the homogeneous coordinates of this point work out to be $(b^2\lambda/\rho^2,b\tau,b\mu-\tau)$. Substituting the coefficients from (3) and dehomogenizing yields $$-{\cos\theta \over R\cos\theta+b\sin\theta}\left(b^2/\rho^2,Rb\right).\tag5$$

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  • $\begingroup$ Thanks for the answer! I don't understand the notation in the final formula. Is the coefficient supposed to be multiplied by each of the elements of the point (b^2/p^2, Rb)? That would give me the point (S, H)? $\endgroup$ – Orion DeYoe Oct 31 '19 at 21:00
  • $\begingroup$ @OrionDeYoe That’s right. I factored out a clumsy common term. $\endgroup$ – amd Oct 31 '19 at 22:16

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