Question

Identify and trace the conic x²-2xy+y²-3x+2y+3=0

Answer 1

$$x^2 - 2xy + y^2 - 3x + 2y + 3 = 0 \text{ } (1)$$

We give a quadratic form:

$$x^2 - 2xy + y^2$$

to the canonical form

$$B= \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}$$

Find the eigenvalues and eigenvectors of this matrix:

$$(1 - λ)x_1 -1y_1 = 0 \\ -2x_1 + (1 - λ)y_1 = 0$$

Characteristic equation:

$$\begin{vmatrix} 1 - λ & -1 \\ -1 & 1 - λ \end{vmatrix}= λ^2- 2λ = 0$$

$$λ^2 -2 λ = 0 \\ D=(-2)^2 - 4*1=4$$

Find the main axes of the quadratic form, that is, the eigenvectors of the matrix B.

$$λ_1 = 0 \\ x_1-y_1 = 0 \\ -x_1 + y_1 = 0$$

 or

$$x_1-y_1 = 0$$

The eigenvector $(λ_1 = 0 , for x_1 = 1)$

: 

$$\vert \bar{x_1}\vert=\sqrt{1^2+1^2}=\sqrt{2}$$

So

The coordinates of the eigenvector (eigenvalue $λ_2 = 2$):

$$-x_1-y_1 = 0 \\ -x_1-y_1 = 0$$

or

$$-x_1-y_1 = 0$$

The eigenvector

$$(λ_2 = 2, for x_1=1)$$

$$\vert \bar{x_2}\vert=\sqrt{1^2+(-)1^2}=\sqrt{2}$$

Moving on to a new basis:

From (1) obtain

$$({\frac{1}{\sqrt{2}} x_1+\frac{1}{\sqrt{2}} y_1})^2 - 2({\frac{1}{\sqrt{2}} x_1+\frac{1}{\sqrt{2}} y_1})({\frac{1}{\sqrt{2}} x_1-\frac{1}{\sqrt{2}} y_1}) +\\ + ({\frac{1}{\sqrt{2}} x_1-\frac{1}{\sqrt{2}} y_1})^2 - 3({\frac{1}{\sqrt{2}} x_1+\frac{1}{\sqrt{2}} y_1}) + 2({\frac{1}{\sqrt{2}} x_1-\frac{1}{\sqrt{2}} y_1}) + 3 = 0$$

Now, completing the squares, we get

$$( y_1- \frac{5}{4} \frac{1}{\sqrt{2}})^2 = \frac{1}{2} (\frac{1}{\sqrt{2}}x_1- \frac{23}{16})$$

Or

$$( y_1- \frac{5}{4\sqrt{2}} )^2 = 2\frac{1}{4}\frac{1}{\sqrt{2} }(x_1- \frac{23}{16}\sqrt{2})$$

We received the equation of a parabola:

 $(y - y0)^2 = 2p(x - x0) , p=\frac{1}{4\sqrt{2}}$