Distribution of $\frac{X_1X_3+X_2X_4}{X_3^2+X_4^2}$ 
$X_1, X_2, X_3$ and $X_4$ are independent standard normal random variables. Find the distribution of $$T=\frac{X_1X_3+X_2X_4}{X_3^2+X_4^2}$$

I have found that $U=X_1X_3+X_2X_4$ follows standard Laplace distribution. And $V=X_3^2+X_4^2$, being the sum of squares of two independent standard normal random variables, follows Chi-square distribution with d.f. $2$.
But I don't know whether $U$ and $V$ are independent. Because I only have the marginal distributions of $U$ and $V$(i.e. standard Laplace and Chi-square, respectively), but not the joint distribution of $(U,V)$. So I cannot check whether they are independent by checking if their joint distribution is product of their marginal distributions. How can I know that $U$ and $V$ are independent, since both have common terms like $X_3$ and $X_4$? Is there any other way to check their independence?
Since I am not sure about their independence, I cannot proceed to find the distribution of $\frac{U}{V}$ by using Jacobian technique, because for that joint distribution of $(U,V)$ is necessary.
Please anyone help me clear this doubt. Thanks in advance. 
 A: Suppose $$Z=\frac{X_1X_3+X_2X_4}{\sqrt{X_3^2+X_4^2}}$$
Now conditioning $X_3=a,X_4=b$, we find that the above has a standard normal distribution, since
$$\frac{aX_1+bX_2}{\sqrt{a^2+b^2}}\sim N(0,1)$$
for any real $a,b$. As the conditional distribution does not depend on $a,b$, the unconditional distribution remains the same. That is, $Z\sim N(0,1)$.
Relating $Z$ and $T$ we have $$T=\frac{Z}{\sqrt{X_3^2+X_4^2}}=\frac{Z}{\sqrt 2\sqrt{(X_3^2+X_4^2)/2}}$$
As the distribution of $Z$ is independent of $X_3,X_4$, we can see that $$\frac{Z}{\sqrt{(X_3^2+X_4^2)/2}}\sim t_2$$
So I think $T$ is distributed as $\frac1{\sqrt 2}$ times a $t$ distributed variable with $2$ degrees of freedom.
A: Write $W = (X_1,X_2)$ and $Z = (X_3,X_4)$, which are i.i.d random vectors (with standard normal coordinates). Then using your notation, $U = W\cdot Z$ and $V = \|Z\|^2$ so $T= (W\cdot Z)/\|Z\|^2$. So using that $W$ and $Z$ are independent and law of total probability we have for any Borel $A$ that
\begin{align*}
P(T\in A) &= \int P\left(\frac{W\cdot z}{\|z\|^2}\in A\mid Z = z\right)P(Z=z)dz \\
&=
\int P\left(\frac{W\cdot z}{\|z\|^2}\in A\right)P(Z=z)dz.
\end{align*}
Since you know the distributions of $W$ and $Z$ individually you should be able to work out the distribution. (I haven't myself checked it so let me know if there is any difficulties).
A: The numerator and denominator of your random variable $T$ are, in fact, independent, since their parameters do not depend on each other. One of the ways to find the distribution of $T$ is by conditioning on $X_3, X_4$ as follows $$X_1X_3 + X_2X_4|X_3, X_4\sim \mathcal{N}(0, X_3^2 + X_4^2)$$ a.s. and hence $$\frac{X_1X_3 + X_2X_4}{\sqrt{X_3^2+ X_4^2}}|X_3, X_4\sim \mathcal{N}(0, 1).$$ Since the value of the term does not depend on the values of $X_3, X_4$, we can conclude that it is independent of $X_3$ and $X_4$. Moreover, we can conclude that the distribution of $T$ is a product of independent standard normal and a random variable $\frac{1}{\sqrt{W}}$ with $W$ being chi-squared with $df=2$, which is, in fact, a t-distribution with $df = 2$ multiplied by a  constant $\frac{1}{\sqrt{2}}$.
