Start out by graphing $y=x^2$, and labeling lines $\overline{AB}$ and $\overline{CD}$, such that all points $A,B,C,D$ are on the line $y=x^2$. We label the point $A (a,a^2)$, the point $B (b,b^2)$, the point $C (c,c^2)$, and the point $D(d,d^2)$. The lines intersect at the y-axis, which we label to be point $(0,e)$. We start by finding the equation $\overline{AB}$ is on. We put the line into the slope-intercept form $y=mx+z$, where $m$ is the slope and $z$ is the y-intercept* . Since the line intersects the $y$-axis at the point $(0,e)$, we know the y-intercept is $e$. The slope of the line is $\frac{b^2-a^2}{b-a}=\frac{(b+a)(b-a)}{(b-a)}=b+a$. Therefore we have $y=(a+b)\times x+e$. Substituting the point $(a,a^2)$ into the equation gives us $a^2=(a+b)\times a+e\implies a^2=a^2+ab+e\implies ab+e=0$. We could also substitute $(b,b^2)$ into the equation to give us $b^2=(a+b)\times b+e\implies ab+e=0$ as well. We now find the equation $\overline{CD}$ is on. Again, put the line into slope-intercept form $y=mx+z$, where $m$ is slope, and $z$ is y-intercept*. The lines $\overline{AB}$ and $\overline{CD}$ intersect at the point $(0,e)\forall e$, therefore the y-intercept of $\overline{CD}$ is $e$. The slope of $\overline{CD}$ is $\frac{d^2-c^2}{d-c}=\frac{\left(d+c\right)\left(d-c\right)}{\left(d-c\right)}= d+c$. Therefore our new equation is $y= \left(d+c\right)\times x+e$. Substituting either points $(c,c^2)$ or $(d,d^2)$ give $c^2= \left(d+c\right)\times c+e$ or $d^2= \left(d+c\right)\times d+e$, where both can be simplified two after expanding and subtract $c^2$ in the first equation, and $d^2$ in the second equation to give $cd+e=0$. We are now left with the two equations $cd+e=0$ and $ab+e=0$. Subtracting the two equations results in \[cd-ab=0\] \[\implies cd=ab\] \[\implies d=\boxed{\frac{ab}{c}}\] *(note: I changed the slope-intercept form from $y=mx+b$ to $y=mx+z$ to avoid any confusion with point $B$)