Find constant $k$ for the following multivariate normal distribution Suppose $X,Y$ are random variables that have the following joint probability density function:$$f(x,y) = k \cdot \text{exp}\left\{-\frac{1}{2}\left[\frac{4}{3}x^2 +\frac{16}{3}y^2 + \frac{8}{3}xy -8x -16y +16\right]\right\}$$ 
find the appropriate $k$. 
Since this exercise corresponds to a multivariate normal distribution section I assume the function must match the following: https://en.wikipedia.org/wiki/Multivariate_normal_distribution.
I have tried matching my given equation with the bivariate case, but I just seem to be getting huge and tedious equations and I don't think this is what my professor had in mind. 
Is there some property or shortcut that could be useful to find the constant that I'm missing?
NOTE: the exercise then follows asking you to find $\sigma_i, \mu_i \ldots$ which is why I believe I should match the functions!
Thanks for the help!
 A: Equate power of exponents:
$$\tag{*}\label{*}
\frac{4}{3}x^2 +\frac{16}{3}y^2 + \frac{8}{3}xy -8x -16y +16=\frac{(x-\mu_X)^2}{(1-\rho^2)\sigma^2_X}+\frac{(y-\mu_Y)^2}{(1-\rho^2)\sigma^2_Y}-\frac{2\rho(x-\mu_X)(y-\mu_Y)}{(1-\rho^2)\sigma_X\sigma_Y}$$
Then equate coefficient under $x^2$, $y^2$, $xy$:
$$\tag{1}\label{1}
\frac{1}{(1-\rho^2)\sigma^2_X}=\frac43,\quad \frac{1}{(1-\rho^2)\sigma^2_Y}=\frac{16}3, \quad \frac{2\rho}{(1-\rho^2)\sigma_X\sigma_Y}=-\frac83.
$$
First two equations
$$\tag{2}\label{2}
(1-\rho^2)\sigma^2_X=\frac34,\quad (1-\rho^2)\sigma^2_Y=\frac{3}{16}
$$
imply that $\sigma_X=2\sigma_Y$. Substitute this equality into the third equation in (\ref{1}) give:
$$
\frac{2\rho}{(1-\rho^2)2\sigma^2_Y}=\frac{\rho}{(1-\rho^2)\sigma^2_Y}=-\frac83.
$$
Substituting denominator of this fraction by $\frac{3}{16}$ from (\ref{2}) leads to $\rho=-\frac12$. 
After that obtain variances from (\ref{2}): $\sigma_X=1$, $\sigma_Y=\frac12$.
After that I see the real need to check whether the values that we obtained bring the equality into the very first equation (\ref{*}). In other words, it is equuvalent to checking whether given p.d.f. is really p.d.f. of multivariate normal. But instead of tedious calculus you can simply put l.h.s. into Wolfram Alpha and see that iso-density loci are ellipses. 
Finding $\mu_X=2$ and $\mu_Y=1$ became a simple taks after you put $\rho$ and $\sigma_X,\sigma_Y$ into the r.h.s. of (\ref{*}) and equate few coefficients under $x$ and $y$.
