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Similar to how we can get an eqn. of a line by dotting a vector between two points with it's perpendicular vector

i.e: $(a,b) \cdot (x-x_o,y-y_o)$=$ax+by-ax_o-by_o$

David K
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tryst with freedom
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1 Answers1

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There are related questions such as Conic Sections with Matrices in which we find out that the vector equations of conics are typically put into the form something like

$$ \newcommand{x}{\mathbf x}\newcommand{xhat}{\hat{\mathbf{ x}}}\newcommand{L}{\mathbf h}\newcommand{T}{^{\sf T}} \newcommand{p}{\mathbf p} \x\T A \x - 2\L\T\x + f = 0, \tag1$$

where $\x$ and $\L$ are $2\times1$ vectors, $A$ is a $2\times2$ matrix, and $f$ is a scalar. There are several variations on this depending on how you want to define and label the constant parts, but the basic idea is the sum of a quadratic form in $\x,$ a linear map $\mathbb R^2 \to \mathbb R$ over $\x,$ and a scalar.

I strongly suspect this is not a complete answer yet because you might like to see how to get such an equation from a geometric definition of a conic Now, much as your example produced the equation of a line from the fact that two points determine a line, we can use geometric facts to derive the equations of the conic sections This is what I have time for at the moment, but I may want to expand on it.

Let's try a parabola. To keep the equations simpler, assume the $y$-axis is the directrix. Then the distance from the directrix to a point $\x$ in the half plane $x > 0$ is $\xhat\T \x$ where $\xhat = (1,0)\T$ is the unit vector in the positive $x$ direction. The square of the distance on either side of the $y$ axis is therefore $$(\xhat\T \x)^2 = (\x\T \xhat)(\xhat\T \x) = \x\T \begin{pmatrix} 1&0 \\ 0&0 \end{pmatrix} \x. $$

Let $\p$ (a point with $x>0$) be the focus of the parabola. Then the square of the distance from $\p$ to $\x$ is $$(\x - \p)\T (\x - \p) = \x\T\x - 2\p\T\x + \p\T\p.$$

The equation of a parabola on the right side of the $y$ axis should set the squares of the two distances equal, so we have $$ \x\T\x - 2\p\T\x + \p\T\p = \x\T \begin{pmatrix} 1&0 \\ 0&0 \end{pmatrix} \x,\tag2$$ which is equivalent to $\x\T A \x + 2\L\T\x + f = 0$ (Equation $(1)$) where $$A = \begin{pmatrix} 0&0 \\ 0&1 \end{pmatrix}, \quad \L = \p, \quad \text{and}\quad f = \p\T \p.$$ For example, if $\p = (2,0)\T$ this equation works out to $y^2 - 4x + 4 = 0.$

But an ellipse or hyperbola also has a directrix (two of them, in fact), and we can generalize Equation $(2)$ to $$ \x\T\x - 2\p\T\x + \p\T\p = \rho^2 \x\T \begin{pmatrix} 1&0 \\ 0&0 \end{pmatrix} \x. \tag3$$

Equation $(3)$ is equivalent to Equation $(1)$ with $$A = \begin{pmatrix} 1 - \rho^2 &0 \\ 0&1 \end{pmatrix}, \quad \L = \p, \quad \text{and}\quad f = \p\T \p.$$ If $\p = (2,0)\T$ this works out to $y^2 + (1 - \rho^2)x^2 - 4x + 4 = 0,$ which is an ellipse when $\rho < 1,$ a parabola when $\rho = 1,$ and a hyperbola when $\rho > 1.$

A circle of radius $r$ about $\p$ of course is given by $$(\x - \p)\T (\x - \p) = \x\T\x - 2\p\T\x + \p\T\p = r^2,$$ which is equivalent to Equation $(1)$ with $$A = I, \quad \L = \p, \quad \text{and}\quad f = \p\T \p - r^2.$$

Of course equations can also be derived for a directrix that might be at any angle and might not pass through the origin. There are just more parameters to deal with in the general case.

There's another form of the vector equations of a conic in which we use three-dimensional vectors instead of two-dimensional vectors. Then the general equation is $$ \x\T A \x = 0 $$ where $A$ is a $3\times3$ matrix. You can get the two-dimensional coordinates of a conic by restricting $\x$ to vectors of the form $(x, y, 1)\T.$ For example, see the question how to simplify a general plane conic section's equation by linear algebra? The two forms of the equation are related to each other as explained in an answer to Projective conic section.

David K
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