(b) The ellipse $E$ has equation $\mathbf X^T\mathbf A\mathbf X=24$ where $\mathbf X=\binom xy$ and $\mathbf A$ is defined as in part (a).
(i) Write down the equation of $E$ in terms of $x$ and $y$.
(ii) Show that $E$ can be rotated about the origin onto the ellipse $E'$ having equation $2x^2+3y^2=6$.
From this problem, how is one able to solve part B. For part A, the matrix is given to be $$\mathbf A=\begin{pmatrix}11&\sqrt3\\\sqrt3&9\end{pmatrix}$$ So I found the eigenvalues and eigenvectors of the this matrix to be $12$ and $8$, and the eigenvectors to be $\langle 3,\sqrt3\rangle$ and $\langle-1,\sqrt3\rangle$. Additionally solving part 1 of question b, gives $11x^2+2xy\sqrt{3}+9y^2=24$. How does one show that this can be rotated about the origin to get to $2x^2+3y^2=6$? I used desmos to visualize this and using slopes I figured out it was $60^\circ$.
$E$ may also be written as $\mathbf X^T(\mathbf A/4)\mathbf X=6$, i.e. $$\mathbf A/4=\begin{bmatrix}11/4&\sqrt3/4\\\sqrt3/4&9/4\end{bmatrix}$$ This has eigenvalues $3$ and $2$. Now the matrix formulation of $2x^2+3y^2=6$ is $\mathbf X^T\mathbf B\mathbf X=6$ where $$\mathbf B=\begin{bmatrix}2&0\\0&3\end{bmatrix},$$ and $\mathbf B$ also has eigenvalues $3$ and $2$. This tells us that $\mathbf A/4$ and $\mathbf B$ are similar. Since these matrices are symmetric, there exists an orthogonal $\mathbf P$ such that $\mathbf P^T(\mathbf A/4)\mathbf P=\mathbf B$, and the geometric interpretation of $\mathbf P$ is a rotation of the ellipse around the origin. This proves part (b).