Determine conic section $x^2-4xy+4y^2-6x-8y+5=0$ and its center So I got a task to determine the conic section of the following:
$$x^2-4xy+4y^2-6x-8y+5=0$$
I started using matrices and got to the equation :
$$5x`^2 + \frac {10x`}{\sqrt5}-\frac {20y`}{\sqrt5}+5=0$$
I completed the square and got to the equation :
$$\frac {\sqrt5(x`+\frac {1}{\sqrt5})^2}{4} + \frac {1}{\sqrt5}=y`$$
So I can see it's most likely a parabola.
But I'm confused - How can I surely know ? How can I find the "center" (tasked asked for center but probably meant vertex).
I know that somehow I need to find $x`$ and $y`$ and then multiply it by my orthogonal $P$ matrix and this will be the center.
 A: For large $x$ and $y$, the equation becomes
$$x^2-4xy+4y^2=(x-2y)^2=0$$
indicating a parabola, with the symmetry line $y=\frac12x$. Let the tangential line to the vertex  be $y=-2x+b$ and substitute it into the conic $x^2-4xy+4y^2-6x-8y+5=0$ to get
$$25x^2+10(1-2b)x+4b^2-8b+5=0$$
Its discriminant has to be zero, yielding $b=1$, and the equation reduces to
$$25x^2-10x+1=(5x-1)^2=0$$
Then, solve for the vertex $(\frac15,\frac35)$.
A: The given equation of the conic is
$ x^2 - 4 x y + 4 y^2 -6x - 8y + 5 = 0 $
If we define $r = [x, y]^T $ then the given conic equation can be written in matrix-vector form as
$ r^T A x + b^T x + c = 0 $
where
$ A = \begin{bmatrix} 1 && - 2 \\ -2 && 4 \end{bmatrix} $
$ b = [-6, -8] $
$ c = 5
First we need to check if $A$ is invertible.  Calculate the determinant
$ | A | = (1)(4) - (-2)^2 = 0 $
So $A$ is singular, therefore the conic is either a parabola or a pair of lines.
Diagonalize $A$ and put it in the form $A = R D R^T $, using the following steps.
Step 1:  Calculate the angle $\phi = \dfrac{1}{2} \tan^{-1}\bigg( \dfrac{2 A_{12}}{A_{11} - A_{22} } \bigg) = \dfrac{1}{2} \tan^{-1}\big(\dfrac{-4}{-3}\big) $
It follows that $ \tan(2 \phi) = \dfrac{4}{3} \Longrightarrow \cos(2 \phi) = \dfrac{3}{5} , \sin(2 \phi) = \dfrac{4}{5} $
Step 2:  It follows from step 1. that $\cos(\phi) = \sqrt{ \dfrac{(1 + \cos(2 \phi) }{2} } = \dfrac{2}{\sqrt{5}} $ and $\sin(\phi) = \sqrt{ \dfrac{(1 - \cos(2\phi) }{2} } = \dfrac{1}{\sqrt{5}} $
Step 3:Define the rotation matrix as
$R = \begin{bmatrix} \cos(\phi) && - \sin(\phi) \\ \sin(\phi) && \cos(\phi) \end{bmatrix} = \dfrac{1}{\sqrt{5}} \begin{bmatrix} 2 && - 1 \\ 1 && 2 \end{bmatrix}$
Step 4: Compute the diagonal elements of matrix $D$
$D_{11} = \dfrac{1}{2}(A_{11} + A_{22}) + \dfrac{1}{2} (A_{11} - A_{22}) \cos(2 \phi) + A_{12} \sin(2 \phi)  = \dfrac{5}{2} - \dfrac{9}{10} -  \dfrac{8}{5} = 0 $
$D_{22} = \dfrac{1}{2}(A_{11}+A_{22}) - \dfrac{1}{2}(A_{11} - A_{22}) \cos(2 \phi) - A_{12} \sin(2 \phi) = \dfrac{5}{2} + \dfrac{9}{10} + \dfrac{8}{5} = 5 $
Step 5:  Now the equation of the conic is
$ r^T R D R^T r + b^T r + c = 0 $
So define $w = [w_1, w_2] =  R^T r \Longleftrightarrow r = R w $, then
$ w^T  D w + b^T R w + c = 0$
where $b^T R = \dfrac{1}{\sqrt{5}} \begin{bmatrix} -20 , -10 \end{bmatrix} $
This equation written explicitly is
$ 5 w_2^2 - \dfrac{20}{\sqrt{5}} w_1 - \dfrac{10}{\sqrt{5}} w_2 + 5 = 0 $
Dividing by $5$, it becomes
$ w_2^2 - \dfrac{4}{\sqrt{5}} w_1 - \dfrac{2}{\sqrt{5}} w_2 + 1 = 0 $
Completing the square in $w_2$:
$ (w_2 - \dfrac{1}{\sqrt{5}})^2 - \dfrac{4}{\sqrt{5}} w_1 + \dfrac{4}{5} = 0 $
Thus, $w_1 = \dfrac{\sqrt{5}}{4} \bigg( (w_2 - \dfrac{1}{\sqrt{5}})^2 + \dfrac{4}{5} \bigg)  $
Which is clearly a parabola that has a vertex in the $w$-plane equal to $w =(\dfrac{1}{\sqrt{5}}, \dfrac{1}{\sqrt{5}})$
The vertex in the $xy$ plane is found by the relation $ r = R w $.  Using this, we get
$ \text{Vertex} = \frac{1}{5} \begin{bmatrix} 1 \\ 3 \end{bmatrix} = ( \dfrac{1}{5}, \dfrac{3}{5})$
