Hello, Rimas!
A line is tangent to a parabola $\displaystyle y^2=4ax$ at a point $\displaystyle P$ (not the vertex)
and intersects the axis of this parabola at a point $\displaystyle Q.$
Prove that the line segment $\displaystyle PQ$ is bisected by the line
which is tangent to the parabola at its vertex.
If they made a sketch, they could have stated the problem more clearly. Code:

 P o *
 o *
 o* (p,q)
o *
o *
o 
o      *         
Q 
*
 *
 *

The parabola is "horizontal", vertex at (0,0), opens to the right.
The tangent at $\displaystyle P(p,q)$ has its xintercept at $\displaystyle Q.$
Prove that $\displaystyle PQ$ is bisected by the yaxis.
We have: .$\displaystyle y^2 = 4ax \quad\Rightarrow\quad 2yy' = 4a \quad\Rightarrow\quad y' = \frac{2a}{y}$
At $\displaystyle P(p,q)$, the slope of the tangent is: .$\displaystyle m = \frac{2a}{q}$
The equation of the tangent is: .$\displaystyle y  q \:=\:\frac{2a}{q}(xp) \quad\Rightarrow\quad y \:=\:\frac{2a}{q}x + \frac{q^22ap}{q}$
Since $\displaystyle q^2 = 4ap$, we have: .$\displaystyle y \:=\:\frac{2a}{y}x + \frac{4ap  2ap}{q} \quad\Rightarrow\quad y \:=\:\frac{2a}{q}x + \frac{2ap}{q} $
Let $\displaystyle y = 0$, the xintercept is: .$\displaystyle Q(p,0)$
Let $\displaystyle x = 0$, the yintercept is: .$\displaystyle M\left(0, \frac{2ap}{q}\right)$ .[1]
Since $\displaystyle q^2 = 4ap$, then $\displaystyle 2ap = \frac{q^2}{2}$
. . Substitute into [1]: .$\displaystyle M\left(0, \frac{\frac{q^2}{2}}{q}\right) \:=\:M\left(0,\frac{q}{2}\right) $
And $\displaystyle M\left(0,\frac{q}{2}\right)$ is indeed the midpoint of $\displaystyle P(p,q)$ and $\displaystyle Q(p,0)$