User:Egm6322.s09.Three.ge/MyReport7

=Canonical Form of Ellipse=

HW:Deriving the Canonical Form for an Ellipse

Given the general,modified form of a PDE: $$\lambda_{1}\bar{x}^2+\lambda_{2}\bar{y}^2+d\bar{x}+e\bar{y}+f=0 \qquad \ldots (1)$$

Derive the canonical form representing an ellipse from (1).

The canonical form representing an ellipse is given by the following equation

$$\xi^{2}+ \eta^2 = g \qquad \ldots (2)$$

where $$\xi$$ and $$\eta$$ represent axes in coordinate system that coincide with the principle axes of the ellipse.

In order to show equation (1) in the form of equation (2)the coordinate system $$(\bar{x},\bar{y})$$ must be translated and then rotated.

Step(1)- Translation

Define $$\begin{bmatrix} \bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}= \begin{bmatrix}\bar{x}+r \\ \bar{y}+s\end{bmatrix}$$

Substituting $$\bar{\bar{x}}$$ and $$\bar{\bar{y}}$$ in (1) yields:

$$\lambda_{1}(\bar{\bar{x}}-r)^2+\lambda_{2}(\bar{\bar{y}}-s)^2+d(\bar{\bar{x}}-r)+e(\bar{\bar{y}}-s)+f=0$$

Expanding the squared terms gives

$$\begin{matrix} \Rightarrow \lambda_{1}\bar{\bar{x}}^{2}+\lambda_{2}\bar{\bar{y}}^2+\bar{\bar{x}}(d-2\lambda_{1}r)+\bar{\bar{y}}(e-2\lambda_{2}s)+ f=0 \\ \therefore \\ r= \frac{d}{2\lambda_{1}} \qquad s= \frac{e}{2\lambda_{2}} \end{matrix}\qquad \ldots (3)$$

Where the values for $$r$$ and $$s$$ are set so that equation (3) resembles (2). Specifically, $$r$$ and $$s$$ are found by solving the following equations.

$$\begin{matrix}(d-2\lambda_{1}r)=0 \\ \\ (e-2\lambda_{2}s)=0 \end{matrix}$$

Substituting $$r$$ and $$s$$ into equation (3), the equation reduces to:

$$\begin{matrix}\lambda_{1}\bar{\bar{x}}^2+\lambda_{2}\bar{\bar{y}}^2-\frac{d}{4\lambda_{1}^{2}}-\frac{e}{4\lambda_{2}^{2}}+f=0 \\ \textrm{Let}\quad g=\left[-\frac{d}{4\lambda_{1}^{2}}-\frac{e}{4\lambda_{2}^{2}}+f \right ] \\ \end{matrix}$$

Thus,

$$ \lambda_{1}\bar{\bar{x}}^2+\lambda_{2}\bar{\bar{y}}^2=g \quad\ldots (4)$$

Step(2)- Rotation

Equation (4) can be rewritten as $$\left \lfloor \bar{\bar{x}} \ \bar{\bar{y}} \right \rfloor

\begin{bmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \\ \end{bmatrix}

\begin{Bmatrix} \bar{\bar{x}}\\ \bar{\bar{y}} \end{Bmatrix}=g \quad \ldots (5)$$

Now one applies rotation of coordinates by defining to arrive at the new coordinate system $$(\xi, \eta)$$.

Define the rotation as:

$$\begin{bmatrix} \xi \\ \eta \end{bmatrix}=

\mathbf{J_{\beta}} \begin{bmatrix} \bar{\bar{x}}\\ \bar{\bar{y}} \end{bmatrix}$$

where $$\mathbf{J_{\beta}}= \begin{bmatrix} \sqrt{\frac{g}{\lambda_{1}}} & 0 \\ 0 & \sqrt{\frac{g}{\lambda_{2}}} \end{bmatrix}$$

Since $$\mathbf{J_{\beta}}$$ being an orthogonal matrix, (i.e. $$\mathbf{J_{\beta}}^{-1}= \mathbf{J_{\beta}}^{T}= \mathbf{J_{\beta}}$$) it can be shown that:

$$\begin{bmatrix} \bar{\bar{x}}\\ \bar{\bar{y}} \end{bmatrix}=

\mathbf{J_{\beta}}^{-1} \begin{bmatrix}\xi \\ \eta \end{bmatrix}=

\mathbf{J_{\beta}} \begin{bmatrix}\xi \\ \eta \end{bmatrix}$$

Substituting this rotation of coordinates into equation (5) results in:

$$\left \lfloor \xi \quad \eta \right \rfloor

\mathbf{J_{\beta}} \begin{bmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \end{bmatrix}

\mathbf{J_{\beta}} \begin{Bmatrix} \xi \\ \eta \end{Bmatrix} =g$$

Plugging in the value of $$\mathbf{J_{\beta}}$$ yields the following expression.

$$\left\lfloor \xi \quad \eta \right\rfloor

\begin{bmatrix} \sqrt{\frac{g}{\lambda_{1}}} & 0 \\ 0 & \sqrt{\frac{g}{\lambda_{2}}} \end{bmatrix}

\begin{bmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \end{bmatrix}

\begin{bmatrix} \sqrt{\frac{g}{\lambda_{1}}} & 0 \\ 0 & \sqrt{\frac{g}{\lambda_{2}}} \end{bmatrix}

\begin{Bmatrix} \xi \\ \eta \end{Bmatrix}

=g$$

Simplifying the above equation further, we obtain

$$g\xi^{2}+ g\eta^{2} = g$$

Which clearly reduces to:

$$\xi^{2}+ \eta^{2} = 1$$