3. [6 points] In the plane $O x y$, the equation $26 a^{2}-22 a x-20 a y+5 x^{2}+8 x y+4 y^{2}=0$ defines the coordinates of point $A$, and the equation $a x^{2}+2 a^{2} x-a y+a^{3}+1=0$ 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 $3 x-y=4$ (points $A$ and $B$ do not lie on this line).
Answer: $(-\infty ;-1) \cup\left(-\frac{1}{3} ; 0\right) \cup\left(\frac{8}{3} ;+\infty\right)$.

Solution. The first equation can be transformed as follows:

$$
\begin{aligned}
& \left(4 y^{2}+4(2 x-5 a) y\right)+5 x^{2}-22 a x+26 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left((2 y)^{2}+2 \cdot 2 y \cdot(2 x-5 a)+(2 x-5 a)^{2}\right)-(2 x-5 a)^{2}+5 x^{2}-22 a x+26 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow(2 y+2 x-5 a)^{2}+x^{2}-2 a x+a^{2}=0 \Leftrightarrow(2 y+2 x-5 a)^{2}+(x-a)^{2}=0
\end{aligned}
$$

From this, we obtain that the coordinates of point $A\left(x_{A} ; y_{A}\right)$ satisfy the equations $2 x+2 y-5 a=0$ and $x-a=0$, i.e., $x_{A}=a, y_{A}=\frac{3 a}{2}$.

Since the second equation defines a parabola, $a \neq 0$. Under this condition, the equation can be rewritten as $y=x^{2}+2 a x+\left(a^{2}+\frac{1}{a}\right)$. The coordinates of point $B$ are: $x_{B}=-a, y_{B}=y(-a)=\frac{1}{a}$.

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 a \in(-\infty ;-1) \cup\left(-\frac{1}{3} ; 0\right) \cup\left(\frac{8}{3} ;+\infty\right)
\end{aligned}
$$