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I'm looking for a proof of this following theorem :

Given $ax^2+bxy+cy^2=k$, to standardize equation of this conic section with coefficient of $x'$ being $a'$ and coefficient of $y'$ being $b'$, then $a'$ and $b'$ are the solutions to this equation:

$$m^2-(a+c)m+\left( ac-\frac{b^2}{4} \right) = 0$$

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    See https://en.wikipedia.org/wiki/Rotation_of_axes#Rotation_of_conic_sections – 2017-02-03

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I would do it like this

$\mathbf x^T \begin{bmatrix} a & \frac b2 \\ \frac b2 & c\end{bmatrix} \mathbf x = 0$

I say that there exist vectors such $Av = \lambda v$

If such vectors exist, then: $Av = \lambda I v\\ (A-\lambda I)v = 0$

$(A-\lambda I)$ is a singular matrix.

$\det(A-\lambda I) = 0$

$\det\begin{bmatrix} a-\lambda & \frac b2 \\ \frac b2 & c-\lambda\end{bmatrix} = \lambda^2 - (a+c)\lambda + ac - \frac {b^2}{4} = 0$

lets call the roots of that polynomial $\lambda_1, \lambda_2$

$A v_1 = \lambda_1 v_1,A v_2 = \lambda_1 v_2\\ A[v_1,v_2] = [v_1,v_2] \begin{bmatrix} \lambda_1 &\\&\lambda_2\end{bmatrix}$

$AP = P D\\ A = PDP^{-1}$

I am not going to prove it here but if $A$ is symmetric,

1) it is diagonalizable

2) it has real eigenvalues

3) there is an ortho-normal set of eigenvectors

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    I've seen the Matrix but I want to prove it – 2017-02-03
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    Sorry that is the proof. $x^T A x = x^TP^T D P x \implies P$ is your rotation, the elements of $D$ are the coefficients of your rotated conic, and the elements of $D$ are the roots of that polynomial. – 2017-02-03