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 $(y - a)^2 + x^2 + \frac{5}{4}a x - \frac{5}{4}x y = 0$.

Since each point at which these lines intersect hyperbola satisfies $x y = 1$, the latter equation for them can be written as $x^2 + \frac{5}{4}a x + (y - a)^2 - \frac{5}{4} = 0$. 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 $y = -1.6x$.