User:Egm6322.s09.bit.la.hw6

from p 19.2: $$\mathbf{\bar{A}}=\mathbf{J}\mathbf{A}\mathbf{J}^T$$

$$det \mathbf{\bar{A}}=det\mathbf{J}*det\mathbf{A}*det(\mathbf{J})^T $$

where: $$det(\mathbf{J})^T= det\mathbf{J}$$

$$det\mathbf{\bar{A}}=(det\mathbf{A})(det\mathbf{J})^2$$

$$\bar{a}\bar{c}-\bar{b}^2=(ac-b^2)(\phi _x \psi_y- \phi _y \psi x)^2$$

Equations for conics in 2D

$$\lfloor x y\rfloor\begin{bmatrix} a & b\\ b & c\end{bmatrix}\begin{Bmatrix}x\\ y\end{Bmatrix}+\lfloor d e \rfloor\begin{Bmatrix}x\\ y\end{Bmatrix} + f =0$$

$$ ax^2+2bxy+cy^{2}+dx+ey+f=0$$

Figure

This are called canonical forms because they generate circles, ellipses, parabolas, hyberpolas by cutting the cone with a plane in different sections

Canonical forms

Ellipses

$$\left ( \frac{x}{a} \right )^2+\left ( \frac{y}{b} \right )^2=1$$

where : $$\left ( \frac{x}{a} \right )=\xi $$

and $$\left ( \frac{y}{b} \right )=\eta$$

If in the above equation a=b then you will have a circle (specific case of ellipses)

Parabolas

$$\pm \xi^2-\eta = 0$$

when:

$$+\xi \Rightarrow$$ concave in $$+\eta$$ direction

$$-\xi \Rightarrow$$ concave in $$-\eta$$ direction

$$ax^2-y=0$$

$$\pm (\sqrt{\left | a \right |}x)^2-y=0$$

Where:

$$\sqrt{\left | a \right |}x=\xi $$

and $$y=\eta$$

Hyperbolas

$$\xi ^2-\eta ^2=\pm 1 $$

$$\xi\eta=\pm 1$$

Goal: Find coordinate transformation that removes mixed products xy

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Explanation of $$\mathbf{J}^{-1}=\mathbf{A}$$

$$\mathbf{J}\mathbf{J}^{-1}=\mathit{I}$$

$$ \begin{bmatrix}cos\theta & sin\theta \\ -\frac{sin\theta }{r}&\frac{cos\theta }{r} \end{bmatrix}\begin{bmatrix} a & b\\c & d\end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$

Solving the system below to find $$\mathbf{J}^{-1}$$

$$cos\theta a + sin\theta c=1$$

$$cos\theta b + sin\theta d=0$$

$$\frac{ -sin\theta}{r}a+\frac{ cos\theta}{r}c=0$$

$$\frac{ -sin\theta}{r}b+\frac{ cos\theta}{r}d=1$$

Puting a,b,c and d in the matrix form:

$$\mathbf{J}^{-1}=\begin{bmatrix} a & b\\c & d\end{bmatrix}=\begin{bmatrix} cos\theta &-rsin\theta  \\ sin\theta & rcos\theta \end{bmatrix}=\mathbf{A}$$

Showing that $$\mathbf{J}^{-1}=\mathbf{A}$$