By shifting the origin to the point $(2,3)$ and
then rotating the coordinate axes through an angle $\theta$ in counter clockwise direction, the
$x=2+X \cos \theta-Y \sin \theta$
and $y=3+x \sin \theta+y \cos \theta$, so the given equation $3 x^2+2 x y+3 y^2-18 x-22 y+50=0$ is transformed to
$$
\begin{aligned}
& 3(2+x \cos \theta-y \sin \theta)^2+2(2+x \cos \theta-y \sin \theta) \\
& (3+x \sin \theta+y \cos \theta) \\
& +3(3+x \sin \theta+y \cos \theta)^2-18(2+x \cos \theta \\
& -y \sin \theta)-22(3+x \sin \theta+y \cos \theta)+50=0
\end{aligned}
and given as $4 x^2+2 y^2-1=0$ and as $x y$ is missing so $\tan 2 \theta=\frac{2}{3-3} \Rightarrow 2 \theta=\frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4}$