9. Find all values of the parameter $a$ for which the system of equations $\left\{\begin{array}{c}4 \sqrt{y}=x-a, \\ y^{2}-x^{2}+2 y-4 x-3=0 .\end{array}\right.$ has a unique solution. Specify this solution for each $a$.
# Solution:

$\left\{\begin{array}{c}4 \sqrt{y}=x-a, \\ y^{2}-x^{2}+2 y-4 x-3=0 .\end{array}\right.$ Transform the second equation of the system.

$\left(y^{2}+2 y+1\right)-\left(x^{2}+4 x+4\right)=0,(y+1)^{2}-(x+2)^{2}=0,(y+x+3)(y-x-1)=0$.

1) $y+4 \sqrt{y}+a+3=0, \sqrt{y}=-2 \mp \sqrt{4-a-3}=-2 \mp \sqrt{1-a}$. One root, $\sqrt{y}=-2-\sqrt{1-a}<0,$ is extraneous, the other, $\sqrt{y}=-2+\sqrt{1-a} \geq 0$, if $\sqrt{1-a} \geq 2$, i.e., when $a \leq-3$.
2) $y-4 \sqrt{y}-a-1=0, \sqrt{y}=2 \pm \sqrt{4+a+1}=2 \pm \sqrt{a+5}$.

$(\sqrt{y})_{1}=2+\sqrt{a+5} \geq 0$, if $a+5 \geq 0, a \geq-5 .(\sqrt{y})_{2}=2-\sqrt{a+5} \geq 0$, if

$\sqrt{a+5} \leq 2,0 \leq a+5 \leq 4,-5 \leq a \leq-1$. When $a=-5(\sqrt{y})_{1}=(\sqrt{y})_{2}=2$

![](https://cdn.mathpix.com/cropped/2024_05_06_450936e33dcef50d2d99g-04.jpg?height=205&width=1325&top_left_y=1488&top_left_x=434)

Comparing the solutions from 1) and 2), we see that the system of equations has a unique solution when $a \in(-\infty ;-5) \cup(-1 ;+\infty)$.

Answer: $a \in(-\infty ;-5), x=4(-2+\sqrt{1-a})+a, y=(-2+\sqrt{1-a})^{2}$;

$$
a \in(-1 ;+\infty), x=4(2+\sqrt{a+5})+a, y=(2+\sqrt{a+5})^{2}
$$