Answer to Question #91515 in Analytic Geometry for Ra

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:

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