Process of finding the canonical form of conic section

Question

I have a problem with understanding the process of transforming a conic section into canonical form.

My conic is:

$11x^2-24xy+4y^2+2x+16y-11=0$

Can someone explain me the complete process?

Answer 1

Option 1.

Do it with trig.

$x = u \cos\theta - v \sin \theta\\ y = u \sin \theta + v\cos \theta\\ x^2 = u^2\cos^2\theta + v^2\sin^2\theta -2uv\sin 2\theta\\ y^2 = u^2\cos^2\theta + v^2\sin^2\theta +2uv\sin 2\theta\\ xy = (u^2-v^2)\sin\theta\cos\theta + uv\cos 2\theta$

We are going to make the substitutions above and find $\theta$ such that the coefficients of the uv terms equal 0.

$-7\sin 2\theta - 24\cos 2\theta = 0\\ \tan 2\theta = -\frac {24}7\\ \cos 2\theta = \frac {7}{\sqrt{7^2+24^2}} = \frac {7}{25}\\ \cos\theta = \sqrt {\frac {1+\frac 7{25}}{2}} = \frac{4}{5}\\ \sin\theta = -\frac {3}{5}$

$(11\cdot \frac{16}{25} + 4\cdot \frac9{25} + 24\cdot\frac{24}{25})u^2 + (11\cdot \frac{9}{25} + 4\cdot \frac{16}{25} - 24\cdot\frac{24}{25})v^2+(2\cdot\frac 45 - 16\cdot\frac 35)u + (2\cdot\frac 35 + 16\cdot\frac 45) v = 11\\ 20u^2 - 5v^2 -8u+14v=11$

And you can get home from there.

option 2: factor

$11x^2 -24 xy + 4 y^2 +2x + 16 y - 11 = 0\\ (11 x - 2y -9 )(x - 2y +1) = 2$

indicates

$11 x - 2y -9 = 0,x - 2y +1=0$

are the asymptotes.

and finally by linear algebra.

$\mathbf {x^T}\begin{bmatrix}11 &- 12\\-12&4\end{bmatrix}\mathbf {x} + \begin{bmatrix}2&16\end{bmatrix}\mathbf {x} = 11$

Diagonalize the matrix on the left.

It has characteristic equation:

$\lambda^2 - 15\lambda + 100 = 0\\ (\lambda -20)(\lambda + 5)$

$\begin{bmatrix}-9 &- 12\\-12&-16\end{bmatrix}$

$\begin{bmatrix} 4\\-3\end {bmatrix}$ is an eigenvector.

Since the matrix is symmetric the other eigenvector will be orthogonal.

$\begin{bmatrix}11 &- 12\\-12&4\end{bmatrix} = \begin{bmatrix} \frac 45&\frac 35\\-\frac 35&\frac 45\end{bmatrix}\begin{bmatrix} 20\\&-5\end{bmatrix}\begin{bmatrix} \frac 45&-\frac 35\\\frac 35&\frac 45\end{bmatrix}$

$\mathbf u = \begin{bmatrix} \frac 45&-\frac 35\\\frac 35&\frac 45\end{bmatrix}\mathbf x$

$\mathbf {u^T}\begin{bmatrix}20 &\\&-5\end{bmatrix}\mathbf {u} + \begin{bmatrix}-8&14\end{bmatrix}\mathbf {u} = 11$

Which is an equivalent expression to the first approach.

Answer 2

In maxima, first find the center

solve([diff(11*x^2-24*x*y+4*y^2+2*x+16*y-11,x)=0,diff(11*x^2-24*x*y+4*y^2+2*x+16*y-11,y)=0],[x,y]);
(%o1) [[x = 1, y = 1]]

Then transform the equation into one centered on the origin:

expand(11*(x+1)^2-24*(x+1)*(y+1)+4*(y+1)^2+2*(x+1)+16*(y+1)-11);
(%02) 4*y^2-24*x*y+11*x^2-2

Resulting in $11(x-1)^2-24(x-1)(y-1)+4(y-1)^2=2$ or the translated $$11x'^2-24x'y'+4y'^2=2.$$

Now we rotate:

$M=\begin{pmatrix}11&-12\\-12&4\end{pmatrix}$

eigenvectors(M);
                                             4           3
(%o4)            [[[- 5, 20], [1, 1]], [[[1, -]], [[1, - -]]]]
                                             3           4

So, normalising the eigenvectors, $P=\begin{pmatrix}3/5&4/5\\4/5&-3/5\end{pmatrix}$ should work:

transpose(P).M.P;
                                  [ - 5  0  ]
(%o6)                             [         ]
                                  [  0   20 ]

So $-5u^2+20v^2=2$ or $$-\frac52 u^2+10 v^2=1.$$