Given a conic $\Gamma$ that has the equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, $\Gamma$ can be represented by the symmetric matrix
$$\mathbf{C} = \begin{bmatrix}
A & B/2 & D/2\\
B/2 & C & E/2\\
D/2 & E/2 & F
\end{bmatrix}.$$
Suppose we have points $(x_\text{c}, y_\text{c})$, $(x_-, y_-)$, $(x_+, y_+)$ and $(x_0, y_0)$ that satisfy the following:
- $\Gamma$ has center $(x_\text{c}, y_\text{c})$.
- $(x_-, y_-), (x_+, y_+) \in \Gamma$.
- The tangents at $(x_-, y_-)$ $(x_+, y_+)$ intersect at $(x_0, y_0)$.
Are these conditions enough to make $\Gamma$ unique, assuming that $\Gamma$ exists? I have two positional and two tangential conditions; does knowledge of the conic center count as the "fifth condition" for specifying conics?
I attempt to solve the equations formed by following this Wikipedia article.
From (1) we have
$$\begin{cases}
Ax_\text{c} + B(y_\text{c}/2) + D/2 = 0\\
B(x_\text{c}/2) + Cy_\text{c} + E/2 = 0.
\end{cases}$$
(2) implies that
$$\left\{\begin{split}
Ax_-^2 + Bx_-y_- + Cy_-^2 + Dx_- + Ey_- + F &= 0\\
Ax_+^2 + Bx_+y_+ + Cy_+^2 + Dx_+ + Ey_+ + F &= 0.
\end{split}\right.$$
Finally, (3) gives us
$$\begin{gather}
\begin{bmatrix}
x_0 & y_0 & 1
\end{bmatrix}\mathbf{C}
\begin{bmatrix}
x_- \\ y_- \\ 1
\end{bmatrix} = 0 =
\begin{bmatrix}
x_0 & y_0 & 1
\end{bmatrix}\mathbf{C}
\begin{bmatrix}
x_+ \\ y_+ \\ 1
\end{bmatrix}\\
\implies
\begin{cases}
Ax_0x_- + B(x_0y_- + y_0x_-)/2 + Cy_0y_- + D(x_0 + x_-)/2 + E(y_0 + y_-)/2 + F = 0\\
Ax_0x_+ + B(x_0y_+ + y_0x_+)/2 + Cy_0y_+ + D(x_0 + x_+)/2 + E(y_0 + y_+)/2 + F = 0\end{cases}
\end{gather}$$
Reorganizing the system of linear equations gives us
$$
\begin{bmatrix}
x_\text{c} & \frac{y_\text{c}}{2} & 0 & \frac{1}{2} & 0 & 0\\
0 & \frac{x_\text{c}}{2} & y_\text{c} & 0 & \frac{1}{2} & 0\\
x_-^2 & x_-y_- & y_-^2 & x_- & y_- & 1\\
x_+^2 & x_+y_+ & y_+^2 & x_+ & y_+ & 1\\
x_0x_- & \frac{x_0y_- + y_0x_-}{2} & y_0y_- & \frac{x_0 + x_-}{2} & \frac{y_0 + y_-}{2} & 1\\
x_0x_+ & \frac{x_0y_+ + y_0x_+}{2} & y_0y_+ & \frac{x_0 + x_+}{2} & \frac{y_0 + y_+}{2} & 1
\end{bmatrix}
\begin{bmatrix}
A \\ B \\ C \\ D \\ E \\ F
\end{bmatrix} = \mathbf{0}
$$
However, this means that the square matrix on the left-hand-side must be singular. How do I go about solving for one permissible solution of $\begin{bmatrix}A & B & C & D & E & F\end{bmatrix}^\intercal$? The solution vectors should all be parallel.
Best Answer
Hmmm. Still stuck for a counterexample. Since nobody answered, I'll hazard some initial thoughts.
1) The general solution should always be singular.
2) First, we need definitions sufficient to distinguish a solution.
Have you worked out the uniqueness condition for any individual cases? For example, consider a parabola with axis $ O + \lambda \vec x_1$, O on the Directrix. The center is $(x_c, y_c) =O\pm \infty \vec x_1.$
Definitions needed.
3) Without additional information, I'd write out the general solution and plug in values I already know. What conditions resolve the scalar parameter(s)?
4) Break it up into subproblems. E.g.
i) (under what conditions) is the solution unique?
ii) what is the form of the general solution?
iii) How does it change when the origin is translated?.
In the matrix, the parameters $(x_c, y_c)$ complicate the solution. Set the origin to (0,0). All solutions which are unique for C=(0,0) will be unique for arbitrary C.
Formally, translate the coordinate origin to $(h, k)$ by $(\bar x, \bar y) = (x-h, y-k)$, and solve in coordinates $(\bar x, \bar y)$. (The math doesn't seem to mind if I leave the bars off.)
Then let $x_c, y_c$ be free parameters, and ask"How does a translation of origin affect the solution?"
I hope that helps. You might also consider abandoning the matrix. I think the tacit assumption
The condition swap is linear in $A ...F $ , and in some obvious way,
...is false. I've backed up to examining the conics case by case. I'll tackle it in more depth when time permits**
**I'm always available to solve problems together. I am r_k_h_i_r_s_t (at) gmail (Dot) com. Remove underscores. If the solution is not forthcoming, drop me a line.