3. [6 points] On the plane $O x y$, there is a point $A$, the coordinates $(x ; y)$ of which satisfy the equation $5 a^{2}-4 a x+6 a y+x^{2}-2 x y+2 y^{2}=0$, and a circle with center at point $B$, given by the equation $a^{2} x^{2}+$ $a^{2} y^{2}-4 a^{3} x-2 a x+2 a^{2} y+4 a^{4}+1=0$. Find all values of the parameter $a$ for which points $A$ and $B$ lie on opposite sides of the line $x=3$ (points $A$ and $B$ do not lie on this line).
Answer: $\left(0 ; \frac{1}{2}\right) \cup(1 ; 3)$.

Solution. The first equation can be transformed as follows:

$$
\begin{aligned}
& \left(x^{2}-2(2 a+y) x\right)+2 y^{2}+6 a y+5 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(x^{2}-2 \cdot x \cdot(2 a+y)+(2 a+y)^{2}\right)-(2 a+y)^{2}+2 y^{2}+6 a y+5 a^{2}=0 \Leftrightarrow \\
& \quad \Leftrightarrow(x-2 a-y)^{2}+y^{2}+2 a y+a^{2}=0 \Leftrightarrow(x-2 a-y)^{2}+(y+a)^{2}=0
\end{aligned}
$$

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

Since the second equation defines a circle, $a \neq 0$. Under this condition, the equation can be rewritten as $x^{2}+y^{2}-4 a x-\frac{2 x}{a}+2 y+4 a^{2}+\frac{1}{a^{2}}=0$, from which we further get $\left(x^{2}-2 x\left(2 a+\frac{1}{a}\right)+\left(2 a+\frac{1}{a}\right)^{2}\right)+\left(y^{2}+2 y+1\right)+$ $4 a^{2}+\frac{1}{a^{2}}=1+\left(\frac{1}{a}+2 a\right)^{2}$, and finally $\left(x-2 a-\frac{1}{a}\right)^{2}+(y+1)^{2}=5$. Therefore, the coordinates of point $B$ are: $x_{B}=2 a+\frac{1}{a}, y_{B}=-1$.

For points $A$ and $B$ to lie on opposite sides of the line $x=4$, it is necessary and sufficient that their abscissas satisfy the inequality $\left(x_{A}-3\right)\left(x_{B}-3\right)<0$, i.e., $\left(2 a+\frac{1}{a}-3\right)(a-3)<0$. From this, we get $(a-3)\left(\frac{2 a^{2}-3 a+1}{a}\right)<0 \Leftrightarrow \frac{(a-3)(2 a-1)(a-1)}{a}<0$. Solving this inequality using the interval method, we get the answer: $a \in\left(0 ; \frac{1}{2}\right) \cup(1 ; 3)$.

Remark. To solve the problem, it is sufficient to find the abscissas of points $A$ and $B$.