Determine conic section $x^2-4xy+4y^2-6x-8y+5=0$ and its center

Question

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.

Answer 1

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 $(\frac15,\frac35)$.

Answer 2

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})$