Show that the conditional variance of a Gaussian random vector is equal to a constant almost surely. 
Let $(X,Y)$ be a 2-dimensional Gaussian random variable.


(a) Prove that there are constant $a$ and $b$ such that
\begin{align*}
E[Y\,|\,X]=aX+b
\end{align*}


(b) Prove that the conditional variance defined as
\begin{align*}
\text{Var}(Y\,|\,X)=E\big((Y-E[Y\,|\,X])^2\,|\,X\big)
\end{align*}
is equal to a constant almost surely.

I have solved part (a) as follows:
Note that $(X,Y)$ Gaussian $\implies$ that $(Y-aX, X)$ is Gaussian for any $a$, as for any $\alpha, \beta\in\mathbb{R}$ we have
\begin{align*}
\alpha(Y-aX)+\beta X=(-\alpha a +\beta)X+\alpha Y\,\,\text{is a one dimensional normal rv by definition.}
\end{align*}
Now, we choose $a$ such that $Y-aX$ and $X$ are independent, to this end we must have that
\begin{align*}
0&=\text{Cov}(Y-aX, X)\\
&=\text{Cov}(Y, X)-a\text{Cov}(X, X)\\
&=\text{Cov}(Y, X)-a\sigma_X^2\\
&\iff a=\frac{\text{Cov}(Y, X)}{\sigma_X^2}
\end{align*}
So, with this choice of $a$, $Y-aX$ and $X$ are independent. Thus, we have
\begin{align*}
E(Y-aX\,|\,X)=E(Y-aX)=EY-aEX:=b
\end{align*}
But on the other hand,
\begin{align*}
E(Y-aX\,|\,X)=E(Y\,|\,X)-aE(X\,|\,X)=E(Y\,|\,X)-aX
\end{align*}
And thus, $b=E(Y\,|\,X)-aX\implies E(Y\,|\,X)=aX+b$, as we wished to show.
However, I am having trouble showing part $(b)$, I thought it might have something to do with breaking down $Y$ into independent pieces as $Y=(Y-aX)+aX$ but I am having no luck with that, any help here would be greatly appreciated.

Using the $a$ you described which has that property that $Y-aX$ is independent from $X$
$$\begin{align*} Var(Y|X) &= E[(Y-E[Y|X])^2|X]\\
&= E[(Y^2 - aX-b)^2|X]\\ 
&= E[(Y - aX)^2 - 2b(Y-aX) + b^2|X] \\
&= E[(Y-aX)^2|X] - 2b E[(Y-aX)|X] + b^2\end{align*}$$
Since $Y-aX$ is independent of $X$ and since Borel measurable functions such as $t\to t^2$ preserve independent random variables, then $(Y-aX)^2$ is also independent of $X$. Hence $E[(Y-aX)^2|X] = E[(Y-aX)^2]$ and $E[Y-aX|X] = E[Y-aX]$ by properties of the conditional expectation function. Hence $Var(Y|X)$ is a constant a.s.
$$Var(Y|X)  = E[(Y-aX)^2 - 2b(Y-aX) + b^2] = E[(Y-E[Y|X])]$$
