So we need to show that; for given two lines $l_1,l_2$ and two positive real numbers $c_1,c_2$, there exist only (at most) one parabola $y=x^2+px+q$ that cut $l_1,l_2$ into segments length $c_1,c_2$ respectively. We can set the intersection of two lines to be the origin and denote equation of $l_i$ by $y=t_ix$ where $t_1\neq t_2$. We also note that to get the segment length to be $c_i$, we need only to set the difference of $x$-coordinate of the two intersection points between $l_i$ and the parabola to be $d_i=c_icos(\theta_i)$ where $\theta_i$ denote the angle between $l_i$ and $x$-axis. The $x$-coordinate of the intersection is given by the equation $t_ix=x^2+px+q$ or $x^2+(p-t_i)x+q=0$. Set two roots to be $r_{1,i},r_{2,i}$, we get $(r_{1,i}-r_{2,i})^2=(p-t_i)^2-4q\Rightarrow (p-t_i)^2-4q=d_i^2$. So $d_1^2-d_2^2=(t_1-p)^2-(t_2-p)^2\Rightarrow \frac{d_1^2-d_2^2}{t_1-t_2}=t_1+t_2+2p$. This gives us there exist at most one solution $(p,q)$, and so at most one parabola.