How do I find a rotation that brings this equation of a cone to standard axes?

Question

UNDERGRAD LINEAR ALGEBRA~

I'm really bad at this alternate coordinate stuff because I missed the last bit of classes and that's when we did change of bases. I know for a vector in standard basis $\bar{x}$ we can find a representation for the same thing in a base $B$, denoted $[\bar{x}]_B$ by finding $\bar{C}[\bar{x}]_b = \bar{x}$. So far practice problems have been simple manipulations of matrices, so I'm not sure how to approach this problem:

A conic in the xy coordinate system is $5x^2 -2\sqrt{3}xy+7y^2 = 16$. Find a rotation of coordinates that bring it to standard form. What is the matrix of this rotation? What is the cosine of the angle of the rotation? (this part I can probably do after getting started).

Answer 1

A general rotation matrix has the form $$R=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix},$$ while the equation of the conic can be written as $$\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}5&-\sqrt3\\-\sqrt3&7\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=17.$$ Substiuting $(x,y)^T=R^{-1}(x',y')^T$ into the left-hand side of this equation and using $R^{-1}=R^T$ gives $$\begin{bmatrix}x'&y'\end{bmatrix} R\begin{bmatrix}5&-\sqrt3\\-\sqrt3&7\end{bmatrix} R^T \begin{bmatrix}x'\\y'\end{bmatrix} = \begin{bmatrix}x'&y'\end{bmatrix} \begin{bmatrix}6-\cos2\theta+\sqrt3\sin2\theta & -\sqrt3\cos2\theta-\sin2\theta \\ -\sqrt3\cos2\theta-\sin2\theta & 6+\cos2\theta-\sqrt3\sin2\theta\end{bmatrix} \begin{bmatrix}x'\\y'\end{bmatrix}.$$ We want the $x'y'$ term to vanish, so we must have $\sqrt3\cos2\theta+\sin2\theta=0.$ Solve for $\theta$. Note that the rows of the resulting rotation matrix will be eigenvectors of the symmetric matrix associated with the conic, but there’s no need to go through the usual eigenvector/eigenvalue computation to solve this problem.

Answer 2

One straightforward way of doing this is to find a rotation matrix, $\ U\ $, such that the matrix $$ D=U^\top\pmatrix{5&-\sqrt{3}\\-\sqrt{3}&7}U $$ is diagonal. The equation of your conic can be written as $$ \pmatrix{x&y} \pmatrix{5&-\sqrt{3}\\-\sqrt{3}&7} \pmatrix{x\\y}=16\ , $$ so if you define new coordinates $\ x', y'\ $ by $$ \pmatrix{x',y'}=U^{-1} \pmatrix{x\\y}\ , $$ —that is, with respect to axes subjected to the rotation determined by $\ U\ $—then the equation in the new coordinates is \begin{align} 16&= \pmatrix{x'&y'}U^\top \pmatrix{5&-\sqrt{3}\\-\sqrt{3}&7} U\pmatrix{x'\\y'}\\ &= \pmatrix{x'&y'}D \pmatrix{x'\\y'}\\ &= d_1x'^2+d_2y'^2\ , \end{align} where $\ d_1, d_2\ $ are the (diagonal) entries of $\ D\ $.

The columns of $\ U\ $ need to be chosen as normalised eigenvectors of the matrix $\ \pmatrix{5&-\sqrt{3}\\-\sqrt{3}&7}\ $, and $\ d_2, d_2\ $ will then be the corresponding eigenvalues.