9. Find all values of $a$ for which the system of equations

$$
y-2=a(x-4), \quad \frac{2 x}{|y|+y}=\sqrt{x}
$$

has at least one solution, and solve it for each $a$.

Domain of definition: $y>0, x \geq 0$.

In the domain of definition, the second equation of the system takes the form: $x=y \sqrt{x}$.
I. $x=0, y=2-4 a>0$, hence $a<\frac{1}{2}$.
II. $x>0, y=\sqrt{x} ; \sqrt{x}-2=a(\sqrt{x}-2)(\sqrt{x}+2)$.

1) $\sqrt{x}=2, x=4, y=2, a \in \mathbb{R}$.
2) $1=a(\sqrt{x}+2), \sqrt{x}=\frac{1}{a}-2=\frac{1-2 a}{a}>0,0<a<\frac{1}{2}$.

The found solution $x=\left(\frac{1-2 a}{a}\right)^{2}, y=\frac{1-2 a}{a}$ coincides with the previous one if $2=\frac{1}{a}-2, a=\frac{1}{4}$.

Thus, for $a \in\left(0 ; \frac{1}{4}\right) \cup\left(\frac{1}{4} ; \frac{1}{2}\right) \quad x=\left(\frac{1-2 a}{a}\right)^{2}, y=\frac{1-2 a}{a}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_331dee27c7cee369cafag-06.jpg?height=229&width=1373&top_left_y=1553&top_left_x=387)
# Answer:

$$
\begin{aligned}
& a \in(-\infty ; 0] \cup\left\{\frac{1}{4}\right\}, x_{1}=0, y_{1}=2-4 a ; x_{2}=4, y_{2}=2 \\
& a \in\left(0 ; \frac{1}{4}\right) \cup\left(\frac{1}{4} ; \frac{1}{2}\right), x_{1}=0, y_{1}=2-4 a ; x_{2}=4, y_{2}=2 ; x_{3}=\left(\frac{1-2 a}{a}\right)^{2}, y_{3}=\frac{1-2 a}{a} ; \\
& a \in\left[\frac{1}{2} ;+\infty\right), x=4, y=2
\end{aligned}
$$