Question :
The equation $3x^2+2xy+3y^2-18x-22y+50=0$ is transformed to $4x^2+2y^2=1$ with respect to a rectangular axes through the point $(2,3)$ inclined to the original axes at an angle $\theta$. Then $\theta$ is
(a) $\pi/3$
(b) $\pi/8$
(c) $\pi/4$
(d) $3\pi/2$
Progress :
Put $x=x+2$ and $y=y+3$ in the given equation, we get
$3(x+2)^2+2(x+2)(y+3)+3(y+3)^2-18(x+2)-22(y+3)+50=0$
By using a suitable translation---in fact, the given one, though we don't need to know that here---we can find new coordinates, also (mildly abusively) denoted $(x, y)$, for which the linear terms on the l.h.s. of the equation vanish. NB this translation affects neither the quadratic terms nor the angle of rotation, so we may as well ask about the rotation of the ellipse in these coordinates, which thus has equation $$\phantom{(\ast)} \qquad 3 x^2 + 2 x y + 3 y^2 = C \qquad (\ast)$$ for some suitable constant $C$ (one can check that for the $C$ that occurs here, this equation really does define an ellipse, that is, it is neither empty nor a single point).
Hint The equation $(\ast)$ is symmetric in the variables $x, y$, and hence the ellipse it defines is fixed under the reflection $(x, y) \mapsto (y, x)$.