3. [6 points] On the plane $O x y$, the equation $5 a^{2}+12 a x+4 a y+8 x^{2}+8 x y+4 y^{2}=0$ defines the coordinates of point $A$, and the equation $a x^{2}-2 a^{2} x-a y+a^{3}+4=0$ defines a parabola with vertex at point $B$. Find all values of the parameter $a$ for which points $A$ and $B$ lie on opposite sides of the line $y-3 x=4$ (points $A$ and $B$ do not lie on this line).
Answer: $(-\infty ;-2) \cup\left(0 ; \frac{2}{3}\right) \cup\left(\frac{8}{7} ;+\infty\right)$.

Solution. The first equation can be transformed as follows:

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

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

Since the second equation defines a parabola, $a \neq 0$. Under this condition, the equation can be rewritten as $y=x^{2}-2 a x+\left(a^{2}+\frac{4}{a}\right)$. The coordinates of point $B$ are as follows: $x_{B}=a, y_{B}=y(a)=\frac{4}{a}$.

Let $f(x ; y)=3 x-y+4$. Points $A$ and $B$ lie on opposite sides of the line $3 x-y-4=0$ if and only if $f\left(x_{A} ; y_{A}\right)$ and $f\left(x_{B} ; y_{B}\right)$ are numbers of opposite signs. The latter is equivalent to the inequality $f\left(x_{A} ; y_{A}\right) \cdot f\left(x_{B} ; y_{B}\right)0 \Leftrightarrow \\
& \Leftrightarrow \frac{(7 a-8)(3 a-2)(a+2)}{a}>0 \Leftrightarrow a \in(-\infty ;-2) \cup\left(0 ; \frac{2}{3}\right) \cup\left(\frac{8}{7} ;+\infty\right)
\end{aligned}
$$