Answer to Question #91515 in Analytic Geometry for Ra

Question #91515
Identify and trace the conic x²-2xy+y²-3x+2y+3=0
1
Expert's answer
2019-07-11T13:10:10-0400
x22xy+y23x+2y+3=0 (1)x^2 - 2xy + y^2 - 3x + 2y + 3 = 0 \text{ } (1)

We give a quadratic form:


x22xy+y2x^2 - 2xy + y^2

to the canonical form

B=(1111)B= \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}

Find the eigenvalues and eigenvectors of this matrix:


(1λ)x11y1=02x1+(1λ)y1=0(1 - λ)x_1 -1y_1 = 0 \\ -2x_1 + (1 - λ)y_1 = 0

Characteristic equation:


1λ111λ=λ22λ=0\begin{vmatrix} 1 - λ & -1 \\ -1 & 1 - λ \end{vmatrix}= λ^2- 2λ = 0


λ22λ=0D=(2)241=4λ^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.


λ1=0x1y1=0x1+y1=0λ_1 = 0 \\ x_1-y_1 = 0 \\ -x_1 + y_1 = 0

 or

x1y1=0x_1-y_1 = 0


The eigenvector (λ1=0,forx1=1)(λ_1 = 0 , for x_1 = 1)

x1ˉ=12+12=2\vert \bar{x_1}\vert=\sqrt{1^2+1^2}=\sqrt{2}

So



The coordinates of the eigenvector (eigenvalue λ2=2λ_2 = 2):


x1y1=0x1y1=0-x_1-y_1 = 0 \\ -x_1-y_1 = 0

or

x1y1=0-x_1-y_1 = 0

The eigenvector

(λ2=2,forx1=1)(λ_2 = 2, for x_1=1)




x2ˉ=12+()12=2\vert \bar{x_2}\vert=\sqrt{1^2+(-)1^2}=\sqrt{2}





Moving on to a new basis:





From (1) obtain

(12x1+12y1)22(12x1+12y1)(12x112y1)++(12x112y1)23(12x1+12y1)+2(12x112y1)+3=0({\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


(y15412)2=12(12x12316)( y_1- \frac{5}{4} \frac{1}{\sqrt{2}})^2 = \frac{1}{2} (\frac{1}{\sqrt{2}}x_1- \frac{23}{16})

Or


(y1542)2=21412(x123162)( 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:

 (yy0)2=2p(xx0),p=142(y - y0)^2 = 2p(x - x0) , p=\frac{1}{4\sqrt{2}}






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