3. [6 points] In the plane $O x y$, the equation $5 a^{2}-4 a y+8 x^{2}-4 x y+y^{2}+12 a x=0$ defines the coordinates of point $A$, and the equation $a x^{2}-2 a^{2} x-a y+a^{3}+3=0$ defines a parabola with vertex at point $B$. Find all values of the parameter $a$ for which points $A$ and $B$ lie on the same side of the line $2 x-y=5$ (points $A$ and $B$ do not lie on this line).
Answer: $\left(-\frac{5}{2} ;-\frac{1}{2}\right) \cup(0 ; 3)$.

Solution. The first equation can be transformed as follows:

$$
\begin{aligned}
& \left(y^{2}-4(x+a) y\right)+8 x^{2}+12 a x+5 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(\left(y^{2}-2 \cdot y \cdot(2 x+2 a)+(2 x+2 a)^{2}\right)-(2 x+2 a)^{2}+8 x^{2}+12 a x+5 a^{2}=0 \Leftrightarrow\right. \\
& \Leftrightarrow(y-2 x-2 a)^{2}+4 x^{2}+4 a x+a^{2}=0 \Leftrightarrow(y-2 x-2 a)^{2}+(2 x+a)^{2}=0
\end{aligned}
$$

From this, we obtain that the coordinates of point $A\left(x_{A} ; y_{A}\right)$ satisfy the equations $y-2 x-2 a=0$ and $2 x+a=0$, i.e., $x_{A}=-\frac{a}{2}, y_{A}=a$.

Since the second equation defines a parabola, $a \neq 0$. Under this condition, the equation can be rewritten as $y=x^{2}-2 a x+\left(a^{2}+\frac{3}{a}\right)$. The coordinates of point $B$ are as follows: $x_{B}=a, y_{B}=y(a)=\frac{3}{a}$.

Let $f(x ; y)=2 x-y-5$. Points $A$ and $B$ lie on the same side of the line $2 x-y-5=0$ if and only if $f\left(x_{A} ; y_{A}\right)$ and $f\left(x_{B} ; y_{B}\right)$ are of the same sign. The latter is equivalent to the inequality $f\left(x_{A} ; y_{A}\right) \cdot f\left(x_{B} ; y_{B}\right)>0$. We have

$$
\begin{aligned}
& \left(2\left(-\frac{a}{2}\right)-a-5\right)\left(2 a-\frac{3}{a}-5\right)>0 \Leftrightarrow-(2 a+5) \frac{2 a^{2}-5 a-3}{a}<0 \Leftrightarrow \\
& \Leftrightarrow \frac{(2 a+5)(a-3)(2 a+1)}{a}<0 \Leftrightarrow a \in\left(-\frac{5}{2} ;-\frac{1}{2}\right) \cup(0 ; 3)
\end{aligned}
$$