Let $f(x,y)$ be a bivariate normal p.d.f. and let $c$ be a positive constant so that $c < \left(2\pi \sigma_{X} \sigma_{Y}\sqrt{1-\rho^2}\right)^{-1}$. Show that $c=f(x,y)$ defines an ellipse in the $xy$-plane.
Honestly, this question is beyond me, but I tried to give it a shot.
I know one of the equations of an ellipse on a rectangular coordinate system is given by
$$ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \tag1 $$
where the point $(h,k)$ is the center of the ellipse, and width $2a$ and height $2b$.
Futhermore, the p.d.f. of a bivariate normal distribution is given by
$$ f(x,y)=\frac{1}{2\pi \sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp \Bigg\{-\frac{1}{2(1-\rho^2)} \left[\left(\frac{x-\mu_X}{\sigma_X}\right)^2 -2\rho \left(\frac{x-\mu_X}{\sigma_X}\right)\left(\frac{y-\mu_Y}{\sigma_Y}\right)+\left(\frac{y-\mu_Y} {\sigma_Y}\right)^2\right]\Bigg\} \tag2 $$
From here, I can see that $f(x,y)$ depends on $x,y$ in the exponent of the exponential term. I can sort of see that maybe after simplifying a bit, the exponent may take the form of equation $(1)$, but after that, I don't see how I can conclude it defines an ellipse, since we still have $e$ to some exponent.