$5x^2+8xy+5y^2=\mathbf{x}^TA\mathbf{x}= > (S^T\mathbf{x})^TD(S^T\mathbf{x})=1\left(\frac{x-y}{\sqrt{2}}\right)^2+9\left(\frac{x+y}{\sqrt{2}}\right)^2$ Thus, the equation is that of an ellipse, since it is the sum of two squares. It is tempting to simplify this expression by pulling out factors of 2. However, it is important not to do this. The quantities $c_1=\frac{x-y}{\sqrt{2}},\quad c_2=\frac{x+y}{\sqrt{2}}$ have a geometrical meaning. They determine an orthonormal coordinate system on $\mathbb{R}^2$. In other words, they are obtained from the original coordinates by the application of a rotation (and possibly a reflection). Consequently, one may use the $c_1$ and $c_2$ coordinates to make statements about length and angles (particularly length), which would otherwise be more difficult in a different choice of coordinates (by rescaling them, for instance). For example, the maximum distance from the origin on the ellipse ${c_1}^2 + 9{c_2}^2 = > 1$ occurs when $c_2=0$, so at the points $c1=±1$. Similarly, the minimum distance is where $c_2=±1/3$. (Wikipedia, Principal Axis Theorem)

I am not getting the bold parts. Can anyone explain this? I do understand the parts before the bold ones.


Edit: I do know that this is an ellipse. What I am not sure of is what it means by $c_1$ and $c_2$ being used to determine orthonormal coordinates. (I also do know what orthonormal means, as it appears in elementary linear algebra.)

More specifically, what is "original coordinates", and what rotation + reflection is occuring due to $c_1$ and $c_2$? What would be the center?

  • 1
    $\begingroup$ Please get some graph paper and a calculator and draw a very careful picture of $$ 5 x^2 + 8 x y + 5 y^2 = 18.$$ Given a decimal value for $x,$ you can find the two values of $y$ by the quadratic formula, as long as $|x|$ is not too large. The curve is an ellipse. I could explain this very easily in person...I see, this is from wikipedia, you do not necessarily have the background for this. Well, draw the curve. $\endgroup$ – Will Jagy Aug 24 '12 at 4:25

You can start by verifying that the matrix

$$\mathbf A=\begin{pmatrix}5&4\\4&5\end{pmatrix}$$

corresponding to your conic $\mathbf x^\top\mathbf A\mathbf x=5x^2+8xy+5y^2=1$, where $\mathbf x^\top=(x\quad y)^\top$ is the vector of variables possesses the following eigendecomposition:

$$\mathbf A=\begin{pmatrix}5&4\\4&5\end{pmatrix}=\begin{pmatrix}\tfrac1{\sqrt 2}&-\tfrac1{\sqrt 2}\\\tfrac1{\sqrt 2}&\tfrac1{\sqrt 2}\end{pmatrix}\cdot\begin{pmatrix}9&\\&1\end{pmatrix}\cdot\begin{pmatrix}\tfrac1{\sqrt 2}&-\tfrac1{\sqrt 2}\\\tfrac1{\sqrt 2}&\tfrac1{\sqrt 2}\end{pmatrix}^\top=\mathbf S\mathbf D\mathbf S^\top$$

Note that the orthogonal matrix $\mathbf S$ is in fact the $45^\circ$ rotation matrix, and that in fact tells you how you obtain the coordinate system $(c_1,c_2)$ from the starting coordinate system $(x,y)$. Since it's easier to analyze an ellipse for, e.g., its principal axes, when the ellipse's two axes coincide with coordinate axes, you can think of it this way: you start with a sketch of a tilted ellipse, and to ease the task of "looking" at the ellipse, you tilt the paper it's drawn on by $45^\circ$, since it's easier to "look" at when the relevant axes are horizontal and vertical.


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