Matrix of the quadratic form I was solving the following question,

Let $P = \begin{bmatrix}3 & 1\\1 & 3\end{bmatrix}$. Consider the set $S$
  of all vectors $\begin{pmatrix}x\\y\end{pmatrix}$ such that 
  $a^2 + b^2 = 1$ where $\begin{pmatrix}a \\ b \end{pmatrix} = P \begin{pmatrix}x \\y \end{pmatrix}$. Then $S$ is 
  
  
*
  
*a circle of radius $\sqrt{10}$
  
*a circle of radius $\frac{1}{ \sqrt{10}}$
  
*an ellipse with major axis along $\begin{pmatrix}1 \\1 \end{pmatrix}$
  
*an ellipse with minor axis along $\begin{pmatrix}1 \\1 \end{pmatrix}$
  

On using the above conditions I'm getting the following equation
$$ 10x^2 + 10y^2 + 12xy = 1 $$
which is equation of an ellipse, but I'm unable to get it's major and minor axises. I referred the answer given by "Fly by Night" here. He is saying to look at the matrix of quadratic form 
which I'm getting
$$Q = \begin{bmatrix}10 & 6\\6 & 10\end{bmatrix}$$
with eigenvectors $v_1 = (1,1)$ and $v_2 = (-1, 1)$
Now, these two vectors $v_1$ and $v_2$ should be the axis of conic. Now the model matrix which diagonalizes $Q$ is simply the eigen vector matrix, so the model matri $P = \begin{bmatrix}-1 & 1\\1 & 1\end{bmatrix}$. 
So, my question is how is the above procedure giving me angle of rotation of ellipse? And of course answer to my question which by the way if option (4). Quadratic form of matrix was not in my university syllabus so I'm finding it difficult to apply it.
 A: You may want to search for Conics' Classification by means of linear algebra, and/or you can also try what is written in one of the answers in the link you post:
$$\begin{cases}x=X\cos\theta+Y\sin\theta\\{}\\y=-X\sin\theta+Y\cos\theta\end{cases}\;\;\;\;\;\;\text{so substituting we get:}$$
$$10x^2+12xy+10y^2=1\rightarrow 10\left(X^2\cos^2\theta+XY\sin2\theta+Y^2\sin^2\theta+X^2\cos^2\theta-XY\sin2\theta+Y^2\sin^2\theta\right)+$$
$$+12\left((Y^2-X^2)\cos\theta\sin\theta+XY\cos2\theta\right)=1\iff$$
$$20\left(X^2\cos^2\theta+Y^2\sin^2\theta\right)+12\left((Y^2-X^2)\cos\theta\sin\theta+XY\cos2\theta\right)=1$$
Equating the $\;xy\;$ term to zero (as we'd like it to vanish...) we get
$$\cos2\theta=0\iff\theta=\frac\pi4+k\pi\;,\;\;k\in\Bbb Z$$
so we can choose the basic angle $\;\theta=\frac\pi4\;$ , and get the equation:
$$10(X^2+Y^2)+6(Y^2-X^2)=1\iff 4X^2+16Y^2=1\iff\frac{X^2}{\left(\frac12\right)^2}+\frac{Y^2}{\left(\frac14\right)^2}=1$$
and thus this is clearly an ellipse, and in order to get the radiuses of the original ellipse just "go back" rotatinע $\;\frac\pi4\;$ in the clockwise direction, so the minor axis (the vertical one, of length $\;\frac14\;$ , as shown above) alines with the line with a slope of $\;\frac\pi4\;$ with the positive direction of the $\;x\,$ axis, meaning the line $\;y=x\;$ , giving you (4) as the correct option.
