finding the coefficient k of a joint pdf $f(x,y)$ under the multivariate normal distribution given the joint pdf $f(x,y) = $

$$k e^{-\frac12(x^2-2xy+5y^2)},$$

find $k.$

I understand that there is some trick to computing $k$ using properties of the multivariate normal distribution, but have no clue where to start.
$$\int_{-\infty}^\infty e^{-\frac 1 2 (x^2-2xy+5y^2)} \, dx \, dy =\int_{-\infty}^\infty e^{-\frac 1 2 ((x-y)^2+4y^2)} \, dx \, dy. $$

Integrate with respect to $x$ first  and note that

$$
\int_{-\infty}^\infty e^{-\frac 1 2 ((x-y)^2)} \, dx =\int_{-\infty}^\infty e^{-\frac 1 2 x^2} \,dx =\sqrt {2\pi}.
$$

Now $$\int_{-\infty}^\infty e^{-2y^2}=2\sqrt {2\pi}$$ as seen by the substitution $y=2z$. Hence $$\int_{-\infty}^\infty e^{-\frac 1 2 (x^2-2xy+5y^2)} \, dx \, dy =4\pi$$ and the value of the constant $k$ is $\dfrac 1 {4\pi}$.
$$\int_{-\infty}^\infty e^{-\frac 1 2 (x^2-2xy+5y^2)} \, dx \, dy =\int_{-\infty}^\infty e^{-\frac 1 2 ((x-y)^2+4y^2)} \, dx \, dy. $$

Integrate with respect to $x$ first  and note that

$$
\int_{-\infty}^\infty e^{-\frac 1 2 ((x-y)^2)} \, dx =\int_{-\infty}^\infty e^{-\frac 1 2 x^2} \,dx =\sqrt {2\pi}.
$$

Now $$\int_{-\infty}^\infty e^{-2y^2}=2\sqrt {2\pi}$$ as seen by the substitution $y=2z$. Hence $$\int_{-\infty}^\infty e^{-\frac 1 2 (x^2-2xy+5y^2)} \, dx \, dy =4\pi$$ and the value of the constant $k$ is $\dfrac 1 {4\pi}$.