Mean and variance of $Y'\Sigma^{-1}Y-Y_1^2/\sigma_1^2$ when $Y\sim N_2(0,\Sigma)$ 
Let $\underline Y=(Y_1,Y_2)'$ have the bivariate normal distribution $N_2(\underline0,\Sigma)$, where
$$\Sigma=\begin{pmatrix}\sigma_1^2 & \rho\sigma_1\sigma_2 \\[1em] \rho\sigma_1\sigma_2 & \sigma_2^2\end{pmatrix}.\\$$ Obtain the mean and variance of $U=\underline Y'\Sigma^{-1}\underline Y-Y_1^2/\sigma_1^2$.

I could obtain the Mean of $U$ easily since $Y'\Sigma^{-1}Y$ is distributed as Chi- Square with df-2 and $Y_1^2/\sigma_1^2$ with df 1. But I am having trouble obtaining variance since covariance is also present here. Can you please help in finding covariance?
 A: As @whuber mentioned, you have $U=Y'AY$ where $A=\Sigma^{-1}-\begin{pmatrix}1/\sigma_1^2 & 0 \\ 0& 0\end{pmatrix}$.
Note that $A$ is symmetric, so you can use the result in the question here to get the variance:
$$\operatorname{Var}(U)=2\operatorname{tr}((A\Sigma)^2)$$
Also observe that $$A\Sigma=I_2-\begin{pmatrix}1/\sigma_1^2 & 0 \\ 0& 0\end{pmatrix}\Sigma=\begin{pmatrix}0 & -\rho\frac{\sigma_2}{\sigma_1} \\ 0& 1\end{pmatrix}$$
Calculation of the variance would be somewhat simpler if you notice that $A\Sigma$ is an idempotent matrix, which means you can also use this theorem for the exact distribution of $U$:

Suppose $Y\sim N(\mathbf 0,\Sigma)$ where $\Sigma$ is positive definite and let $A$ be a symmetric matrix. Then $Y'AY\sim \chi^2_r$ if and only if $A\Sigma$ is idempotent (or equivalently $A\Sigma A=A$) and $\operatorname{rank}(A\Sigma)=r$.

This is part of a general result proven here. You can find more on distributions of quadratic forms of multivariate normal distribution in standard textbooks (Rao's Linear Statistical Inference and Its Applications and Seber/Lee's Linear Regression Analysis for example).
Indeed $U\sim \chi^2_1$ but that is not to say that difference of arbitrary chi-square variables has a chi-square distribution.
