Given any ellipse in standard form, like
$$x^2/5+y^2/10=1$$ is always elongated towards $y$ as its denominator is larger or $b>a$.
But consider this ellipse $$\frac{\left(x+2y\right)^2}{5}+\frac{\left(2x-y\right)^2}{20}=1$$
This is elongated towards $x+2y=0$, even though it has less denominator.
Whys is that? Can't this ellipse be seen as one with axes $x+2y=0$ and $2x-y=0$
Sketch of this ellipse is here:
The two cases are similar
for $x^2/5+y^2/10=1$ when $x=0$ we have the max elongation towards $y$ (that is $x=0$)
for $\frac{\left(x+2y\right)^2}{5}+\frac{\left(2x-y\right)^2}{20}=1 $ when $x+2y=0$ we have the max elongation towards $2x-y$ (that is $x+2y=0$)
Indeed note that
$$x+2y=0\implies \frac{\left(0\right)^2}{5}+\frac{\left(2x-y\right)^2}{20}=1 \implies \left(2x-y\right)^2=20 \implies25y^2=20\\\implies y=\pm\frac{2\sqrt{5}}{5}\quad x=\mp\frac{4\sqrt{5}}{5}\implies\rho_1=4\frac{\sqrt{6}}5$$
with $\rho_1$ along the $x+2y=0$ axis.
$$2x-y=0\implies \frac{\left(x+2y\right)^2}{5}+\frac{\left(0\right)^2}{20}=1 \implies \left(x+2y\right)^2=5 \implies25x^2=5\\\implies x=\pm\frac{\sqrt{5}}{5}\quad y=\pm \frac{2\sqrt{5}}{5} \implies\rho_2=1$$
with $\rho_2$ along the $2x-y=0$ axis.