Find the equation of an ellipse (isocontour ) I'm a third year student having some difficulties in one of my course because this class is designed for graduate student. 
1.
Let $y=(y_1,y_2)^T\in\mathbb R^2$ a random vector normal distributed with mean 
$\mu = (\mu_1, \mu_2)$ and with variance $\Sigma$, $y\sim N_2(\mu,\Sigma)$.
 Let $U = [u_1;u_2]$ a orthogonal matrix $2\times2$ and $D = \operatorname{diagonal}(\lambda_1,\lambda_2)$, $\lambda_1\gt\lambda_2\gt0$, a diagonal matrix $2\times2$ such that:
$$
\Sigma =
  \begin{bmatrix}
    \sigma_1^2 & \sigma_{12}\\
    \sigma_{12} & \sigma_2^2 
  \end{bmatrix}
= UDU^T$$
Let $w_1 = u_1^Ty$, $w_2 = u_2^Ty$, $\alpha_1 = u_1^T\mu$ and $\alpha_2 = u_2^T\mu$.
2.
The isocontour of a bidimensional normal distribution are ellipses centered in $\mu$. 
We obtain them by: 
$$C = \left\{ y \in \mathbb R^2 : f_y(y)= c \right\} $$
where $f_y(y)$ is the density function of $y$ and $c$ is a positive constant.
Q. Write the equation of the ellipse as function of $c$, $w$, $\lambda_1$, $\lambda_2$, $\alpha_1$ and $\alpha_2$. Show that the set of $C$ describe an ellipse in $\mathbb R^2$. Deduce the principal axis of the ellipse. 
(notes given by my teacher) 
$$ y \in C \Longleftrightarrow f_y(y) = c$$
$$\begin{align} y \in C &\Longleftrightarrow \frac{1}{{(2\pi)}^{\frac{2}{2}}} {\Sigma}^{-\frac12} e^{-\frac12 (y-\mu)^T \Sigma^{-1} (y-\mu)} = c \\
 &\Longleftrightarrow e^{-\frac12 (y-\mu)^T \Sigma^{-1} (y-\mu)} = c \end{align}$$
we are looking at only the points where the density is constant (fixed): 
$$ \Longleftrightarrow e^{-\frac12 (y-\mu)^T \Sigma^{-1} (y-\mu)} = c_1$$
$$ \Longleftrightarrow -\frac12 (y-\mu)^T \Sigma^{-1} (y-\mu) = \log(c_1)$$
where $$c_1 = 2\mu\Sigma^{-\frac12}$$
$$ \Longleftrightarrow (y-\mu)^T \Sigma^{-1} (y-\mu) = \log(c_1) = c_2$$
$$ \Longleftrightarrow (y-\mu)^T \Sigma^{-1} (y-\mu) = -2\log(c_1) = c_3$$
$$ \Longleftrightarrow (y-\mu)^T
  \begin{bmatrix}
    u_1 & u_2
    \end{bmatrix}
  \begin{bmatrix}
    \frac1{\lambda_1}&0\\
    0&\frac1{\lambda_2}\end{bmatrix}
  \begin{bmatrix}
    u_1^T\\
    u_2^T
\end{bmatrix} (y-\mu) =c_3
$$
$$ \Longleftrightarrow
  \begin{bmatrix}
    u_1^T(y-\mu)\\
    u_2^T(y-\mu)\end{bmatrix}^T
  \begin{bmatrix}
    \frac1{\lambda_1}&0\\
    0&\frac1{\lambda_2}\end{bmatrix}
  \begin{bmatrix}
    u_1^T(y-\mu)\\
    u_2^T(y-\mu)\
            \end{bmatrix} =c_3
$$
by 1. 
$w_1 = u_1^Ty$ and $\alpha_1 = u_1^T\mu$
$w_2 = u_2^Ty$ and $\alpha_2 = u_2^T\mu$
$$ \Longleftrightarrow
  \begin{bmatrix}
    w_1-\alpha_1\\
    w_2-\alpha_2\end{bmatrix}^T
  \begin{bmatrix}
    \frac1{\lambda_1}&0\\
    0&\frac1{\lambda_2}\end{bmatrix}
  \begin{bmatrix}
    w_1-\alpha_1\\
    w_2-\alpha_2
\end{bmatrix} =c_3
$$
$$ \Longleftrightarrow \frac{(w_1-\alpha_1)^2}{\lambda_1} + \frac{(w_2-\alpha_2)^2}{\lambda_2} = c_3 $$
I don't understand how did he get from $c$ to $c_3$...
(I get the idea that each $c_1$, $c_2$, $c_3$ represent a contour but I really don't understand how, I don't know how to show that the set of $C$ are ellipse in $\mathbb R^2$)
I'm not sure if the axis are $\pm c\sqrt{\lambda_i}u_i$
any help or hint will be appreciated ! 
 A: This might be easiest to grasp by working backwards from the last equation. Assuming that $c_3\gt0$, we can divide through by it to obtain the equation of an ellipse in standard form: $${(w_1-\alpha_1)^2\over c_3\lambda_1}+{(w_2-\alpha_2)^2\over c_3\lambda_2}=1.$$ This ellipse is centered at $(\alpha_1,\alpha_2)$ and has semi-axis lengths $\sqrt{c_3\lambda_1}$ and $\sqrt{c_3\lambda_2}$. If $c_3=0$, the equation is satisfied only by that center point, which can be considered a degenerate ellipse, while if $c_3\lt0$, the ellipse is imaginary.  
The $w_i$ and $\alpha_i$ in this equation are in a coordinate system that is rotated with respect to the original one. Going back a few steps, we find that this equation is equivalent to $(y-\mu)^T\Sigma^{-1}(y-\mu)=c_3$, which gets expanded into $$\left(U^T(y-\mu)\right)^TD^{-1}\left(U^T(y-\mu)\right)=c_3$$ so this ellipse is centered at $\mu$ in the original coordinate system and its principal axes are $u_1$ and $u_2$.  
As far as the constants $c_1$, $c_2$ and $c_3$ go, they are simply names for more complex expressions that are introduced along the way to reduce clutter. They ultimately derive from the arbitrary constant $c\gt0$ in the definition of the isocontour. For each choice of $c$, you get a different ellipse. You could certainly work backwards through the derivation to get an expression for $c_3$ in terms of the original $c$, but I’m not sure that it would be particularly illuminating. (If you like, you can introduce yet another constant $c_4=\sqrt{c_3}$ so that the half-axes of the ellipse are $c_4\sqrt{\lambda_i}u_i$, but that seems unnecessary to me.)
