Question

Find the orthogonal and Normal canonical reduction of the quadratic form 7x^2 +6xy+7y^2. Hence identify the conic represented by 7x^2+6xy+7y^2=200.. also find the principal axes of the given quadratic form.

Answer 1

Orthogonal reduction

F(x,y)=(x,y) $\begin{bmatrix} 7 & 3 \\ 3 & 7 \end{bmatrix}$ $\begin{pmatrix} x \\ y \end{pmatrix}$ ;

$\begin{vmatrix} 7-k & 3 \\ 3 & 7-k \end{vmatrix}$ =k²-14$\cdot$ k+40=0;

k₁=4, k₂=10;

1)k=4

(3 3)$\cdot\begin{pmatrix} x \\ y \end{pmatrix}$ =$\begin{pmatrix} 0 \\ 0 \end{pmatrix}$

x+y=0;x=-1,y=1;|(-1 1)}=$\sqrt{2}$ ;

v₁=$\frac {1} {\sqrt{2}}\cdot\begin{pmatrix} -1 \\ 1 \end{pmatrix}$ ;

2) k=10

(-3 3)$\cdot\begin{pmatrix} x \\ y \end{pmatrix}$ =$\begin{pmatrix} 0 \\ 0 \end{pmatrix}$

-x+y=0;

x=1,y=1;

v₁=$\frac {1} {\sqrt{2}}\cdot\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ ;

{v₁,v₂}-new ortogonal basis;

P=$\frac {1} {\sqrt{2}}\cdot\begin{pmatrix} 1 &-1 \\ 1 & 1 \end{pmatrix}$

We make substitution

$\begin{pmatrix} \hat{x }\\ \hat{ y } \end{pmatrix}$ =P^-1$\cdot\begin{pmatrix} x \\ y \end{pmatrix}$ =$\frac {1} {\sqrt{2}}\cdot\begin{pmatrix} 1 &1 \\ -1 & 1 \end{pmatrix}$ $\cdot\begin{pmatrix} x \\ y \end{pmatrix}$ $=\frac {1} {\sqrt{2}}\cdot\begin{pmatrix} x+y \\ -x+y \end{pmatrix}$ ;

q(x,y)=($\hat{x} , \hat{y}$ )$\cdot$ $P^{T}\cdot\begin{pmatrix} 7 &3\\ 3 &7 \end{pmatrix}\cdot P\cdot$$\begin{pmatrix} \hat{x }\\ \hat{ y } \end{pmatrix}$=($\hat{x} , \hat{y}$ )$\cdot$$\begin{pmatrix} 10 & 0 \\ 0& 4 \end{pmatrix}$ $\cdot$ $\begin{pmatrix} \hat{x }\\ \hat{ y } \end{pmatrix}$=

=10 $\cdot\hat{x}^{2}+4\cdot\hat{y}^2=200$ ;

${\hat{x}^{2}\over \sqrt{20}^{2}}+{\hat{y}^{2}\over \sqrt{50}^{2}}=1$ it is ellipse.

Principal axes are :

L₁||e₁=$\frac {1} {\sqrt{2}}\cdot\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ ;

L₂||e₂=$\frac {1} {\sqrt{2}}\cdot\begin{pmatrix} -1 \\ 1 \end{pmatrix}$ ;

Canonical reducrion:

7x^2+6xy+7y^2=7$\cdot(x^{2}+\frac {6}{7}\cdot x\cdot y+y^{2})=$

=7$\cdot(x+\frac{3}{7}\cdot y)^{2}+\frac{40}{7}\cdot y^{2}=200$ ;

$\hat{x}=x+\frac{3}{7}\cdot y, \hat{y}=y;$

$7\cdot \hat{x}^2+\frac{40}{7}\cdot \hat{y}^{2}=200;$

${\hat{x}^2\over\sqrt{\frac {200}{7}}^2}+{\hat{y}^2\over\sqrt{35}^2}=1$ - ellipse