The principal axes of the ellipse given by the equation $ 9x^2+6y^2-4xy=4$ are along the directions of the vectors
a) $2i+3j$ and $3i-2j$ b) $2i+j$ and $i-2j$ c) $3i+2j$ and $2i-3j$ d) $i+2j$ and $2i-j$ e) $i+j$ and $i-j$
Seems trivial however I do not have much idea on how to find those vectors.
On
Consider the straight line through the origin
$$y=mx$$ and find its intersections with the ellipse by solving
$$9x^2+6m^2x^2-4mx^2=4,$$
or
$$x^2=\frac4{6m^2-4m+9}.$$
The squared distance from the origin to an intersection is
$$x^2+y^2=x^2(1+m^2)=\frac{4(m^2+1)}{6m^2-4m+9}.$$
By canceling the derivative on $m$, we find the extrema of this function at $$m=2,\\m=-\frac12,$$ corresponding to the direction vectors
$$(1,2),\\(2,-1).$$
The corresponding squared distances are
$$\frac45,\frac25,$$ telling you which are the short and long axis.
Notice that
$$ 9x^2+6y^2-4xy = \begin{pmatrix} x& y \end{pmatrix} \begin{pmatrix} 9 & \frac{-4}{2} \\ \frac{-4}{2} & 6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$ We'll define $\vec{x} = \begin{pmatrix} x \\ y \end{pmatrix} $. Now: $$ 9x^2+6y^2-4xy = \vec{x}^T \begin{pmatrix} 9&-2\\ -2&6 \end{pmatrix} \vec{x} $$ The matrix $\begin{pmatrix} 9&-2\\ -2&6 \end{pmatrix}$ can be diagonalized, which we prefer because then in the new basis there will be no xy term, meaning the new basis will give us the principal axes of the ellipse. Let's find the eigenvalues: $$ (\lambda-9)(\lambda - 6) - 4 = \lambda^2 - 15\lambda + 50 = 0$$ $$ \lambda = 5 ,10 $$ For $\lambda = 5$ we'll find the eigen vector: $$\begin{pmatrix} 4&-2 \\ -2 & 1 \end{pmatrix} v = 0$$ It is easy to see that $v=\begin{pmatrix} 1\\2 \end{pmatrix} $ solves the equation. The original matrix is a symmetric matrix meaning the other eigen vector will be orthogonal to the one we just found meaning it will be $\begin{pmatrix} 2\\-1 \end{pmatrix} $ So the answer is option (d).