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Consider a conic section $C$ in $\mathbb{R}^2$. Every point $P$ in the plane has a "dual" (pole-polar duality) line $L$ with respect to $C$ such that lines $PA$ and $PB$ are tangent to $C$, where $L \cap C = \{A, B\}$. However, given $P, A, B$, is it possible to recover $C$? Is $C$ even uniquely determined by those three points?

My vain attempt is as follows. Suppose $C$ has the matrix representation $\mathbf{C}$, and the images of $P, A, B$ under the map $(x, y) \mapsto \begin{bmatrix}x & y & 1\end{bmatrix}^\intercal$ are $\mathbf{p}, \mathbf{a}, \mathbf{b}$ respectively. Since $A, B \in C$, $\mathbf{a}^\intercal\mathbf{C}\mathbf{a} = 0$ and $\mathbf{b}^\intercal\mathbf{C}\mathbf{b} = 0$. That is, $\mathbf{C}\mathbf{a}$ is orthogonal to $\mathbf{a}$ (note that $(\mathbf{a}^\intercal\mathbf{C})^\intercal = \mathbf{C}\mathbf{a}$ because $\mathbf{C}$ is symmetric). Similarly, $\mathbf{C}\mathbf{b}$ is orthogonal to $\mathbf{b}$.

From tangency we also have $\mathbf{p}^\intercal\mathbf{C}\mathbf{a} = 0$ and $\mathbf{p}^\intercal\mathbf{C}\mathbf{b} = 0$. Since scaling $\mathbf{C}$ doesn't change the fact that it represents $C$, we may set $\mathbf{C}\mathbf{p} = \mathbf{a} \times \mathbf{b}$.

Is it possible to proceed further than this?

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1 Answer 1

up vote 2 down vote accepted

It's not possible to recover C from P, A, B.

One way to see this ... consider A, P, B (in order) to be the control points of a rational quadratic Bezier curve. This curve will be tangent to PA at A, and tangent to PB at B, and it's a conic section curve (all rational quadratic Bezier curves are conics). But, you can adjust the weights of this curve however you please, thereby producing an infinite family of conics.

Another argument ... it's generally well-known that you need 5 positional or tangency constraints to define a conic section curve. You only have four constraints -- two points and two tangent directions.

There's probably an argument based on the idea of "pencils" of conics, too, but two is enough, I hope.

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