Quadratic equation representing a point.

Question

So, we know that $Q(x,y)=Ax^2+Bxy+Cy^2+Dx+Ey+F$ can represent (if not an empty set) an ellipse, a parabola, a hyperbola, a pair of intersecting or parallel lines, a line or a point.

Given the equation in its general form, I'm trying to find out what it represents and I ran into a problem I'm struggling to solve.

How should I proceed to construct an equation that represents a certain point on the plane? For example, the equation $5x^2-2xy+5y^2-4x+20y+20=0$ represents the point $(0,-2)$. Given a point $(x_p,y_p)$ how do I get an equation that represents it?

** For hyperbolas, ellipses, and parabolas a simple rotation followed by a translation will do the trick, for a pair of lines we can just multiply out both of their equations, but what can we do when we want to represent a point?

Would appreciate any kind of help.

Answer 1

$$ Q(x,y) = (x-x_p)^2 + (y-y_p)^2$$

Will work to represent a point by the equation $Q(x,y)=0$.

Mixed terms like $xy$ can be obtained by applying the following kind of linear transformation to your coordinates.

$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} (x-x_p) \\ (y-y_p) \end{bmatrix} $$

With $ad-bc \neq 0$

Answer 2

Let two lines $ax+by+c=0$ and $dx+ey+f=0$ have the intersection $(x_p,y_p)$, then $(ax+by+c)^2+(dx+ey+f)^2=0$ will do the trick.

Answer 3

The canonical degenerate conic that represents a single point is $x^2+y^2=0$. Interpreting this as a degenerate ellipse, we know that its images under an arbitrary invertible affine transformation is another ellipse, with center that’s the image of the original center. So, apply an arbitrary invertible affine transformation that maps the origin to the given point $(x_0,y_0)$ to obtain a conic that represents that point. The simplest such transformation is, of course, a pure translation, yielding $(x-x_0)^2+(y-y_0)^2=0$. Haruhiko Nakahara’s answer is essentially equivalent to this approach: the two intersecting lines there are the images of the coordinate axes under an affine transformation.

Factoring the canonical equation into $(x+iy)(x-iy)=0$ gives us another geometric interpretation: a pair of imaginary lines that intersect in a single real point. In principle, you could pick a pair of imaginary lines that intersect at $(x_0,y_0)$ and multiply their equations, but choosing them so that all of the imaginary terms vanish is much trickier than picking a pair of real lines that pass through the point.