Derivation of the equations for ellipse and ellipsoid Could someone perhaps explain / prove / guide to a source where I can find the derivations of the general equations of ellipse and ellipsoid? I'm trying to understand where these formulas come from:
$$\displaystyle\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
and
$$(\textbf{x}-\textbf{v})^TA^{-1}(\textbf{x}-\textbf{v}) = 1$$
So my question is: "Why these formulas are the way they are?" Proof? Intuition? :) 
Thank you for any help!
 A: For the ellipsoid, just consider the following linear algebra facts.


*

*Symmetric matrices and quadratic forms
Let $A$ be a symmetric real $n\times n$ matrix; then its eigenvalues $\lambda_i$ are real and there exists a set of orthonormal eigenvectors $q_i$ of $A$. We summarize these facts in  matrix form:
$$Q^{t}AQ=\Lambda, $$
or
$$A=Q\Lambda Q^{t}, $$
denoting by $Q$ the orthogonal matrix whose columns are the eigenvectors of $A$ and by $\Lambda$ the diagonal matrix s.t. $\Lambda_{ii}=\lambda_i$. In other words, any real symmetric matrix can be diagonalized.
Now, if the eigenvalues are ordered as follows $\lambda_i\geq \dots\geq \lambda_n$, then
$$x^{t}Ax=x^{t}(Q\Lambda Q^{t})x=(Q^{t}x)\Lambda(Q^{t}x)=(\Lambda~\text{is diagonal})=\sum_{i}=(Q^{t}x)_i \lambda_i (Q^{t}x)_i=\sum_{i}\lambda_i ((Q^{t}x)_i)^2 \leq \lambda_1\|x\|^2.$$
We say that $A$ is positive definite if $x^{t}Ax>0$, for all $x$. This is equivalent to state that all eigenvalues of $A$ are positive.


*

*Ellipsoid


Let $A$ be a positive definite symmetric matrix with real entries. The ellipsoid $\mathcal E_A$ centered at $0$ is the locus 
$$\{x: x^{t}Ax\leq 1 \}. $$
The semi axes are defined as the vectors 
$$s_i=\frac{1}{\sqrt{\lambda_i}}q_i.$$
i.e. the direction of the semi axes are determined by the eigenvectors of $A$, while the eigenvalues $\lambda_i$ determine the "width" of the ellipsoid along the axes.


*

*Easy example


To see the geometry of the ellipsoid definition, let us consider the case $A=diag\{\frac{1}{a^2},\frac{1}{b^2},\frac{1}{c^2} \}$ with $a,b,c>0$. The eigenvalues of $A$ are its diagonal entries. A basis of orthonormal eigenvectors is given by the standard basis $\{e_i\}$ in $\mathbb R^3$ with $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$. Then $Q=Q^t$ and
$$x^{t}Ax=\sum_{i}\lambda_i ((Q^{t}x)_i)^2=\sum_{i}\lambda_i (x_i)^2=
\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}+\frac{x_3^2}{c^2}$$ 
for all $x=(x_1,x_2,x_3)$ in $\mathbb R^3$.
The ellipsoid $\mathcal E_A$ is the locus
$$\{x\in\mathbb R^3: \frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}+\frac{x_3^2}{c^2}\leq 1 \}. $$
The semi axes are (following the above definition)
$$s_1=a(1,0,0), s_2=b(0,1,0), s_3=c(0,0,1).$$
This is coherent with the geometric interpretation of the semi axes in both the 2 and 3-dimensional cases.


*

*And now?


You could produce more complicated examples of ellipsoids by considering non trivial positive definite symm. matrices $A$ and considering ellipsoids not centered at $0$. In this case the definition of the locus is already given in your question.
