$$

Quadratic curve

Problem

Consider the equation

$5x^2-2xy+5y^2=1$. Observe the equation by the basis

$\beta =\{{\mathbf{u}}_1,{\mathbf{u}}_2\}$ such that every point can be written as

$p{\mathbf{u}}_1+q{\mathbf{u}}_2$, where

$${\mathbf{u}}_1=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\end{array}\right]\text{ and }{\mathbf{u}}_2=\frac{1}{\sqrt{2}}\left[\begin{array}{c}-1\\ 1\end{array}\right].$$

What is the equation in terms of

$p$ and

$q$?

Thought

We will convert the equation into a matrix form, called the quadratic form, and then observe the matrix by

$\beta$.

By direct computation, we have

$$a{x}^2+bxy+c{y}^2=1$$

is equivalent to

$$\left[\begin{array}{cc}x& y\end{array}\right]\left[\begin{array}{cc}a& \frac{b}{2}\\ \frac{b}{2}& c\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=1.$$

Thus, the given equation can be written as

$$\left[\begin{array}{cc}x& y\end{array}\right]\left[\begin{array}{cc}5& -1\\ -1& 5\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=1,$$

where the matrix at the center is denoted by

$A$.

Naively, the answer is the equation given by

$$\left[\begin{array}{cc}p& q\end{array}\right][f_A{]}_{\beta }^{\beta }\left[\begin{array}{c}p\\ q\end{array}\right]=1.$$

Here we provide a bit more details. For any point in

${\mathbb{R}}^2$, it can be represented by

$$x{\mathbf{e}}_1+y{\mathbf{e}}_2=p{\mathbf{u}}_1+q{\mathbf{u}}_2,$$

where

${\mathcal{E}}_2=\{{\mathbf{e}}_1,{\mathbf{e}}_2\}$ is the standard basis. Naturally, we have the relations

$$[\mathrm{id}{]}_{{\mathcal{E}}_2}^{\beta }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}p\\ q\end{array}\right]\text{ and }[\mathrm{id}{]}_{\beta }^{{\mathcal{E}}_2}\left[\begin{array}{c}p\\ q\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right].$$

Note that the given basis is orthonormal, so

$[\mathrm{id}{]}_{\beta }^{{\mathcal{E}}_2}$ is an orthogonal matrix, which means

$[\mathrm{id}{]}_{{\mathcal{E}}_2}^{\beta }=([\mathrm{id}{]}_{\beta }^{{\mathcal{E}}_2}{)}^{-1}=([\mathrm{id}{]}_{\beta }^{{\mathcal{E}}_2}{)}^{\mathrm{\top }}$.

By replacing

$x$ and

$y$ by

$p$ and

$q$, we have

$$\left[\begin{array}{cc}x& y\end{array}\right]A\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}p& q\end{array}\right]([\mathrm{id}{]}_{\beta }^{{\mathcal{E}}_2}{)}^{\mathrm{\top }}A[\mathrm{id}{]}_{\beta }^{{\mathcal{E}}_2}\left[\begin{array}{c}p\\ q\end{array}\right]=\left[\begin{array}{cc}p& q\end{array}\right][f_A{]}_{\beta }^{\beta }\left[\begin{array}{c}p\\ q\end{array}\right],$$

leading to the desired result.

Sample answer

We may write the equation

$5x^2-2xy+5y^2=1$ as

$$\left[\begin{array}{cc}x& y\end{array}\right]\left[\begin{array}{cc}5& -1\\ -1& 5\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=1,$$

and denote the central matrix as

$A$. By direct computation, we have

$$\begin{array}{rl}{f}_A({\mathbf{u}}_1)& =\left[\begin{array}{cc}5& -1\\ -1& 5\end{array}\right]\left[\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right]=\left[\begin{array}{c}\frac{4}{\sqrt{2}}\\ \frac{4}{\sqrt{2}}\end{array}\right]=4{\mathbf{u}}_1+0{\mathbf{u}}_2,\\ f_A({\mathbf{u}}_2)& =\left[\begin{array}{cc}5& -1\\ -1& 5\end{array}\right]\left[\begin{array}{c}-\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right]=\left[\begin{array}{c}-\frac{6}{\sqrt{2}}\\ \frac{6}{\sqrt{2}}\end{array}\right]=0{\mathbf{u}}_1+6{\mathbf{u}}_2.\end{array}$$

Consequently,

$$[f_A{]}_{\beta }^{\beta }=\left[\begin{array}{cc}4& 0\\ 0& 6\end{array}\right]$$

and the corresponding equation is

$4p^2+6q^2=1$, which is an equation for an ellipse with semi-major axis of length

$\frac{1}{\sqrt{6}}$ and semi-minor axis of length

$\frac{1}{2}$.

Note

You may use desmos to plot the to equations to observe their relation.

This note can be found at Course website > Learning resources.