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We are given the equation $2x^2 + 4xy + 6y^2 + 6x + 2y = 6$

We did an example of this in class but the equation had less terms.

I took a note in class that says : if there are linear terms, I have to rotate...

This is what I think I have to do.

  1. Put the coefficients of this equation in a matrix $A$, then $B = det(A - I \lambda)$
  2. $det(B)$ must be $ = 0$
  3. Evaluating this determinant will give me a value for lambda that is the root of the equation.
  4. I can then go on and find eigen lines or eigen vectors.

My questions :

  1. Am I right about the steps to solve it ? Shouldn't it give me a canonical form at the end ? because it my notes, it doesn't.

  2. How do I place the coefficients in the matrix A ?

this is my try : $A = \begin{bmatrix}2 & 2\\2 & 6 \\\end{bmatrix}$

the first 2 is for the $2x^2$ term, the second 2 next are for the $4xy$ term which I divided by 2...( I dont really know why btw :\ ) and the 6 is for the $6y^2$ term. Now how do I place the 6x and 2y terms ?

$A = \begin{bmatrix}2 & 2\\2 & 6\\3 & 1 \\\end{bmatrix}$

I add another line with those 2 divided by 2 ?

edit : Here's my current work

I rewrote A as :

$ \begin{bmatrix} x & y \\\end{bmatrix} $ $ A = \begin{bmatrix}2 & 2\\2 & 6 \\\end{bmatrix}$ $ \begin{bmatrix} x \\ y \\\end{bmatrix} $ + $ \begin{bmatrix} 6 & 2 \\\end{bmatrix} $ $ \begin{bmatrix} x \\ y \\\end{bmatrix} $ - 6

$B = \begin{bmatrix}2 - \lambda & 2\\2 & 6 - \lambda \\\end{bmatrix}$

$det(B) = (2 - \lambda)(6 - \lambda) - 4$

$det(B) = 12 - 2 \lambda - 6 \lambda + \lambda^2 - 4 $

And I'm stuck here. How do I factor this ?

$\lambda^2 - 8 \lambda + 8$

edit : so with the quadration formula I found two roots.

$4 + \sqrt{\frac{32}{2}}$ and $4 - \sqrt{\frac{32}{2}}$

Dave
  • 525

3 Answers3

2

Not knowing what exactly "canonical form" is, here is what I get.

Translating to get rid of the linear terms: $$ 2(x+2)^2+4(x+2)(y-1/2)+6(y-1/2)^2=\frac{23}{2}\tag{1} $$ With $P=\dfrac{\sqrt{2+\sqrt2}}{2}\begin{bmatrix}1&1-\sqrt2\\-1+\sqrt2&1\end{bmatrix}=\begin{bmatrix}\cos(\pi/8)&-\sin(\pi/8)\\\sin(\pi/8)&\hphantom{+}\cos(\pi/8)\end{bmatrix}$ we have $$ \begin{bmatrix}2&2\\2&6\end{bmatrix} =P^T \begin{bmatrix}4-\sqrt8&0\\0&4+\sqrt8\end{bmatrix} P\tag{2} $$ Therefore, $$ \begin{bmatrix}x+2\\y-1/2\end{bmatrix}^TP^T \begin{bmatrix}4-\sqrt8&0\\0&4+\sqrt8\end{bmatrix} P\begin{bmatrix}x+2\\y-1/2\end{bmatrix} =\frac{23}2 $$ Thus, the curve is an ellipse with its center at $(-2,1/2)$ and its major axis tilted $\pi/8$ clockwise from the $x$-axis. The major and minor axes are $$ \sqrt{\frac{23}8(2\pm\sqrt2)} $$

To get $(1)$, we translated to get rid of the linear terms. So $$ 2(x+h)^2+4(x+h)(y+k)+6(y+k)^2\\ =(2x^2+4hx+2h^2)+(4xy+4kx+4hy+4hk)+(6y^2+12ky+6k^2)\\ =(2x^2+4xy+6y^2)+(4h+4k)x+(4h+12k)y+(2h^2+4hk+6k^2) $$ To match the linear terms, we need $4h+4k=6$ and $4h+12k=2$. Thus, $h=2$ and $k=-1/2$: $$ 2(x+2)^2+4(x+2)(y-1/2)+6(y-1/2)^2\\ =2x^2+4xy+6y^2+6x+2y+\frac{11}2=6+\frac{11}2=\frac{23}2 $$

To get $(2)$, we diagonalized $\begin{bmatrix}2&2\\2&6\end{bmatrix}$. The matrix $P$ rotates counterclockwise by $\pi/8$. After translating by $(2,-1/2)$ and rotating $\pi/8$ counterclockwise, we are left with the ellipse $$ (4-\sqrt8)x^2+(4+\sqrt8)y^2=\frac{23}2 $$

robjohn
  • 345,667
1

I know of another method, but it may be equivalent to yours: you multiply

$A = \begin{bmatrix}cos\theta & sin\theta\\-sin\theta & cos\theta \\\end{bmatrix}$$ \begin{bmatrix}x\\ y \\\end{bmatrix}$ = $\begin{bmatrix}x' \\y' \\\end{bmatrix}$

And then sub in $(x',y')$ for $(x,y)$ in your original equation, set the $xy$ terms equal to $0$. The solution is the angle $\theta$ by which you must rotate the plane so that the $xy$-terms disappear.

user99680
  • 6,708
0

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{2x^{2} + 4xy + 6y^{2} + 6x + 2y = 6\,,\quad{\large ?}}$

\begin{align} 6&=2x^{2} + 4xy + 6y^{2} + 6x + 2y = \pars{x \quad y} \pars{\begin{array}{cc}2 & 2 \\ 2 & 6\end{array}}{x \choose y} + \pars{3 \quad 1}{x \choose y} + \pars{x \quad y}{3 \choose 1} \\[3mm]&= {\bf v}^{\sf T}A{\bf v} + {\bf a}^{\sf T}{\bf v} + {\bf v}^{\sf T}{\bf a} \qquad\mbox{where}\qquad A \equiv \pars{\begin{array}{cc}2 & 2 \\ 2 & 6\end{array}}\,,\quad {\bf v} \equiv {x \choose y}\,,\quad {\bf a} \equiv {3 \choose 1} \end{align}

$$ {\bf v}^{\sf T}A{\bf v} + {\bf a}^{\sf T}{\bf v} + {\bf v}^{\sf T}{\bf a} = 6 $$ Let ${\bf v} = {\bf r} + {\bf b}$ where ${\bf r}$ represent the 'new coordinates' and ${\bf b}$ is a constant vector which let us to 'remove' the linear term:

\begin{align} 6&=\pars{{\bf r}^{\sf T} + {\bf b}^{\sf T}}A\pars{{\bf r} + {\bf b}}

{\bf a}^{\sf T}\pars{{\bf r} + {\bf b}} + \pars{{\bf r}^{\sf T} + {\bf b}^{\sf T}} {\bf a} = {\bf r}^{\sf T}A{\bf r} + {\bf b}^{\sf T}{\bf b}

  • {\bf r}^{\sf T}\pars{A{\bf b} + {\bf a}}

\[3mm]&+ \pars{{\bf b}^{\sf T}A + {\bf a}^{\sf T}}{\bf r} + {\bf a}^{\sf T}{\bf b}

  • {\bf b}^{\sf T}{\bf a}

\end{align} We choose ${\bf b}$ such that $\quad A{\bf b} + {\bf a} = 0\quad\imp\quad{\bf b} = -A^{-1}a.\quad$ Then,

$$ {\bf r}^{\sf T}A{\bf r}= 6 - {\bf b}^{\sf T}{\bf b} - {\bf a}^{\sf T}{\bf b} - {\bf b}^{\sf T}{\bf a} $$ \begin{align} {\bf b} &= -A^{-1}a =-\,{1 \over 4} \pars{\begin{array}{c}3 & -1 \\ -1 & 1\end{array}}{3 \choose 1} = {-2 \choose 1/2}\,,\qquad \left\lbrace% \begin{array}{rcl} {\bf b}^{\sf T}{\bf b} & = & {17 \over 4} \\ {\bf a}^{\sf T}{\bf b} & = & {\bf b}^{\sf T}{\bf a} = -\,{11 \over 2} \end{array}\right. \end{align} and $\ds{{\bf r}^{\sf T}A{\bf r} = {51 \over 4}}$. With $\ds{{\bf r} = {\bf v} - {\bf b} = {x + 2 \choose y - 1/2}}$ $$ \pars{x + 2 \quad y - \half}\pars{\begin{array}{cc}2 & 2 \\ 2 & 6\end{array}} {x + 2 \choose y - \half} = {51 \over 4} $$

It remains the task of diagonalizing $A$ which amounts to perform a rotation. We hope you can take from here.

Felix Marin
  • 89,464