Find Cross-Tangents to Offset Ellipses

Question

Question:

I am stuying projective geometry and would like to find an analytic solution to my problem. This is a follow-up to my older question.

Given two ellipses with the same major and minor axes, but different origins (written in homogeneous coordinates):

$$ \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1 = 0 $$

and

$$ \frac{(x_2-h)^2}{a^2} + \frac{(y_2-k)^2}{b^2} - 1 = 0 $$

what are the equations of the cross-tangents to the two ellipses? Assume $h, k, a,$ and $b$ are all knowns.

My Attempt:

Following the method from my older question, it makes sense to find the intersection points of the dual curves to each ellipse. Let $f_1$ be the implicit homogeneous function defining our first ellipse (i.e. $f_1(x_1, y_1, z_1) = 0$) and $f_2$ that of our second, respectively. The dual curve of an ellipse is an ellipse (at least, AFAIK, for an ellipse at the origin), which Wikipedia shows nicely.

Hence, our first dual is

$$ a^2 X_1^2 + b^2 Y_1^2 = 1. $$

I am having trouble finding the second dual (namely, that of the ellipse centered at $(h,k)$). Could someone please help me in deriving this? Is this the simplest way to solve this problem?

Following Wikipedia, about the closest I get to is the following and I don't know how to eliminate $\lambda$:

$$ \frac{a^2}{2\lambda} \left( X_2 + \frac{\lambda h}{a^2} \right)^2 + \frac{b^2}{2\lambda} \left( Y_2 + \frac{\lambda k}{b^2} \right)^2 - \left( \frac{\lambda}{2} \right) \left( \left( \frac{h}{a} \right)^2 + \left( \frac{k}{b} \right)^2 \right) + 1 = 0$$

Answer 1

Hint:

Rescale using $au:=x,bv:=y$ and you get two circles,

$$u^2+v^2=1,\\(u-u_c)^2+(v-v_c)^2=1.$$

Rotating around the origin, you can reduce the second to

$$(u-d)^2+v^2=1.$$

The cross-tangents pass through $(\dfrac d2,0)$ and by $(\cos\theta,\sin\theta)$, with

$$\cos\theta\,(\cos\theta-\frac d2)+\sin^2\theta=0.$$

Write the equations of these lines and reverse the transformations.

Answer 2

The dual conics are for $ax^2+bxy+cy^2+dx+ey+f=0$ given by $$ (-\frac14e^2+cf)x^2+(\frac12de-bf)xy+(-\frac14d^2+af)y^2+(-cd+\frac12be)x+(\frac12bd-ae)y-\frac14b^2+ac=0$$ or $\begin{pmatrix}x&y&1\end{pmatrix}\begin{pmatrix}a&\frac{b}{2}&\frac{d}{2}\\\frac{b}{2}&c&\frac{e}{2}\\\frac{d}{2}&\frac{e}{2}&f\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}$ to $$\begin{pmatrix}x&y&1\end{pmatrix}\begin{pmatrix}-\frac14e^2+cf&\frac{\frac12de-bf}{2}&\frac{-cd+\frac12be}{2}\\\frac{\frac12de-bf}{2}&-\frac14d^2+af&\frac{\frac12bd-ae}{2}\\\frac{-cd+\frac12be}{2}&\frac{\frac12bd-ae}{2}&-\frac14b^2+ac\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}$$

from which you can see it's to do with the matrix of a conic and it's adjoint matrix which is the inverse when divided by the determinant. In your special case:

k^2*y^2-b^2*y^2+2*h*k*x*y+2*k*y+h^2*x^2-a^2*x^2+2*h*x+1=0
(-b^2*y^2)-a^2*x^2+1=0

Now solve for intersection points

solve([((b^2*h^2+a^2*k^2)*y^2-h^2)*((b^2*h^2+a^2*k^2)*y^2+4*a^2*k*y+4*a^2-h^2),(2*a^2*b^2*h^3+2*a^4*h*k^2-2*a^2*h^3*k^2)*x+(b^4*h^4*k+2*a^2*b^2*h^2*k^3+a^4*k^5)*y^3+(-b^4*h^4+2*a^2*b^2*h^2*k^2+3*a^4*k^4)*y^2+(2*a^2*b^2*h^2*k-b^2*h^4*k+2*a^4*k^3-3*a^2*h^2*k^3)*y+b^2*h^4-3*a^2*h^2*k^2],[x,y]);
[[x = k/sqrt(a^2k^2+b^2h^2),y = -h/sqrt(a^2k^2+b^2h^2)],
     [x = -k/sqrt(a^2k^2+b^2h^2),y = h/sqrt(a^2k^2+b^2h^2)],
     [x = (ksqrt(a^2k^2+b^2h^2-4a^2b^2)-2b^2h)/(a^2k^2+b^2h^2),
      y = -(hsqrt(a^2k^2+b^2h^2-4a^2b^2)+2a^2k)/(a^2k^2+b^2h^2)],
     [x = -(ksqrt(a^2k^2+b^2h^2-4a^2b^2)+2b^2h)/(a^2k^2+b^2h^2),
      y = (hsqrt(a^2k^2+b^2h^2-4a^2b^2)-2a^2k)/(a^2k^2+b^2h^2)]]

So the cross tangents are $$x \frac{k\sqrt{a^2k^2+b^2h^2-4a^2b^2}-2b^2h}{a^2k^2+b^2h^2}+y \frac{ -(h\sqrt{a^2k^2+b^2h^2-4a^2b^2}+2a^2k)}{a^2k^2+b^2h^2}+1=0\\ x \frac{-(k\sqrt{a^2k^2+b^2h^2-4a^2b^2}+2b^2h)}{a^2k^2+b^2h^2}+ y \frac{h\sqrt{a^2k^2+b^2h^2-4a^2b^2}-2a^2k}{a^2k^2+b^2h^2}+1=0.$$

Answer 3

I wanted to expand on the answer I accepted for my own sake and the sake of those reading this in the future. My post here is also instructive for this problem.

If the equation of a line in line coordintes is

$$ Xx+Yy+1=0 \tag{1} $$

and the tangent to our curve at point $(x_0)$ (identically $(p,q,r)$ on Wikipedia) $<f_x \left. \right|_p, f_y \left. \right|_q, 0>$, denoted $(x')$, is coincident with the line if its inner product with the line coordinates is $0$ per Coxeter

$$ Xx'+Yy'=0 $$

where $(x')=(x',y',0)$, then

$$ Xf_x \left. \right|_p + Yf_y \left. \right|_q = 0. \tag{2} $$

Solving $(1)$ and $(2)$ simultaneously yields

$$ X = \frac{-f_y \left. \right|_q}{xf_y \left. \right|_q-yf_x \left. \right|_p} $$ $$ Y = \frac{f_x \left. \right|_q}{xf_y \left. \right|_q-yf_x \left. \right|_p} $$