69th Belarusian Mathematical Olympiad

b) the required locus is the line $y=-1.6x$.

Let $a$ be the ordinate of the point $T$. Then the lines from the problem condition have equations $y-a-2x=0$ and $y-a-\frac{1}{2}x=0$. The union of these lines is defined by $(y-a-2x)(y-a-\frac{1}{2}x)=0$ which is equivalent to $(y-a{)}^2+x^2+\frac{5}{4}ax-\frac{5}{4}xy=0$.

Since each point at which these lines intersect hyperbola satisfies $xy=1$, the latter equation for them can be written as $x^2+\frac{5}{4}ax+(y-a{)}^2-\frac{5}{4}=0$. This equation defines the circle $$(x+\frac{5}{8}a{)}^2+(y-a{)}^2=\frac{5}{4}+\frac{25}{64}{a}^2,$$ centered at $(-\frac{5}{8}a;a)$. Therefore, all four points of intersection lie on this circle.

Clearly, for any point $T$ on the $y$-axis these lines intersect the hyperbola at exactly four points. Hence the required locus is defined by $\{(-1.6a;a):a\in \mathbb{R}\}$. It is easy to see that the locus is the line $y=-1.6x$.