9. Find all values of the parameter $a$ for which the system of equations $x=4 \sqrt{y}+a, y^{2}-x^{2}+3 y-5 x-4=0$ has a unique solution. Specify this solution for each $a$.

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# Solution:

$\left\{\begin{array}{c}x=4 \sqrt{y}+a, \\ y^{2}-x^{2}+3 y-5 x-4=0 .\end{array}\right.$

Transform the second equation of the system.

$$
\left(y^{2}+3 y+\frac{9}{4}\right)-\left(x^{2}+5 x+\frac{25}{4}\right)=0,\left(y+\frac{3}{2}\right)^{2}-\left(x+\frac{5}{2}\right)^{2}=0,(y+x+4)(y-x-1)=0
$$

1) $y+4 \sqrt{y}+a+4=0, \sqrt{y}=-2 \mp \sqrt{4-a-4}=-2 \mp \sqrt{-a}$. One root, $\sqrt{y}=-2-\sqrt{-a}<0$, is extraneous, the other, $\sqrt{y}=-2+\sqrt{-a} \geq 0$, if $\sqrt{-a} \geq 2$, that is, when $a \leq-4$.
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 $a \geq-5 \cdot(\sqrt{y})_{2}=2-\sqrt{a+5} \geq 0$, 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-09.jpg?height=200&width=1291&top_left_y=1533&top_left_x=491)

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

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