Consider the equation $x^2+xy+3y^2=1$. After an proper counter clock-wise rotation of the coordinate system by an angle $0<\theta<\pi$...

Question

The first part: Consider the equation $x^2+xy+3y^2=1$. After an proper counter clock-wise rotation of the coordinate system by an angle $0<\theta<\pi$, the equation becomes $au^2+bv^2=1$ under the new coordinate system. What is the value of $a+b$?

The second part: What is the angle of the rotation? Round your answer to 2 decimal places.

The approach I am currently taking, is that I know you have to use $\cos(\theta)$, and use a lot of the trig identities. Perhaps it is the trig part that is throwing me off. Plus, it is the rotation of the angle, and whether or not if I have to do any normalizing with this.

Answer 1

The equation may be written as $$ X^T\begin{bmatrix}1&1/2\\1/2&3\end{bmatrix}X=1 $$ where $X=\left[\begin{smallmatrix}x\\y\end{smallmatrix}\right]$. The matrix in the middle is symmetric, and therefore diagonalizable by some orthogonal matrix $P$, giving $$ \begin{bmatrix}1&1/2\\1/2&3\end{bmatrix}=P^TDP $$ (the $P$ and $D$ you will need to calculate, of course). We now have $$ X^TP^TDPX=1\\ (PX)^TD(PX)=1 $$ which we may interpret to mean that $PX=\left[\begin{smallmatrix}u\\v\end{smallmatrix}\right]$. The diagonal entries of $D$ are $a$ and $b$. Also, $P$ (or the matrix you get by swapping the columns of $P$, if $P$ has negative determinant) is the rotation matrix for the rotation you're asked about, so you can simply read off $\cos\theta$ and $\sin\theta$ directly, and calculate $\theta$ from there.

Answer 2

Hint for the first part: Write these equations in matrix form as $\mathbf x^T C \mathbf x = 1$. You can then view the transformation as transforming the original into the diagonal matrix $$\begin{bmatrix}a&0\\0&b\end{bmatrix},$$ and $a+b$ is the sum of its diagonal elements. This sum is called the trace of the matrix, which is unchanged by invertible transformations, so for this problem $a+b=?$

For the second part, yes, you’ll have to work through some trigonometric manipulations, but it shouldn’t be too nasty. You want all of the $xy$ to vanish, so substitute the transformed variables into the original equation, expand and collect up the coefficients of $xy$. That will give you a single equation to solve for $\theta$. The solution is not going to be unique, but all of the solutions will be some multiple of $\frac\pi2$ apart.

Answer 3

I like Arthur's approach, and if you know linear algebra it is the way to go.

If you don't know linear algebra...the alternative is to say

$x = x'\cos \theta + y'\sin\theta\\ y = -x'\sin \theta + y'\cos\theta$

$x^2 = x'^2\cos^2 \theta + y'\sin^2\theta + 2x'y'\sin\theta\cos\theta\\ y^2 = x'^2\sin^2 \theta + y'\cos^2\theta - 2x'y'\sin\theta\cos\theta\\ xy = -x'^2\cos\theta\sin\theta + y'^2\cos\theta\sin\theta + x'y'(\cos^2\theta - \sin^2\theta)$

Plug these into the original equation.

$x'^2(\cos^2\theta + 3\sin^2\theta - \cos\theta\sin\theta) + y'^2(3\cos^2\theta + \sin^2\theta + \cos\theta\sin\theta) + x'y' (-4\sin^2\theta + cos^2\theta - \sin^2\theta) = 1$

Choose $\theta$ such that

$(-4\sin\theta\cos\theta + \cos^2\theta - \sin^2\theta) = 0\\ \cos 2\theta = 2\sin 2\theta\\ \tan 2\theta = \frac 12\\ \theta = \frac 12 \arctan \frac 12$

(and I hope I have rotated the right way...)

$\cos 2\theta = \frac {2}{\sqrt 5}\\ \sin 2\theta = \frac {1}{\sqrt 5}$

$x'^2(-(\cos^2\theta - \sin^2\theta) + 2(\cos^2\theta + \sin^2\theta) - \cos\theta\sin\theta) + y'^2((\cos^2\theta - \sin^2\theta) + 2(\cos^2\theta + \sin^2\theta) + \cos\theta\sin\theta) = 1\\ x'^2(-\cos 2\theta + 2 - \frac 12 \sin 2\theta) + y'^2(\cos 2\theta + 2 + \frac 12 \sin 2\theta) = 1\\ x'^2(-\frac {2}{\sqrt 5} +2 - \frac {1}{2\sqrt 5}) + y'^2(\frac {2}{\sqrt 5} + 2 + \frac 1{2\sqrt 5}) = 1\\ x'^2(2-\frac {\sqrt 5}{2}) + y'^2( 2 + \frac {\sqrt 5}{2}) = 1\\$