The chord joining the points where x = p and x = q on the curve $\mathrm{y} = {\mathrm{ax}}^2 + \mathrm{bx} + \mathrm{c}$ is parallel to the tangent at the point on the curve whose abscissa is

$$\begin{gathered} \mathrm{y}={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{ax}+\mathrm{b} \end{gathered}$$

and points on the curve $\mathrm{A}\left(\mathrm{p}, {\mathrm{ap}}^2+\mathrm{bp}+\mathrm{c}\right) \mathrm{B}\left(\mathrm{q}, {\mathrm{aq}}^2+\mathrm{bq}+\mathrm{c}\right)$

$$\begin{gathered} \therefore \mathrm{slope} \mathrm{of} \mathrm{AB}=\frac{{\mathrm{ap}}^2+\mathrm{bp}+\mathrm{c}-{\mathrm{aq}}^2-\mathrm{bq}-\mathrm{c}}{\mathrm{p}-\mathrm{q}} \\ =\mathrm{a}(\mathrm{p}+\mathrm{q})+\mathrm{b} \\ \frac{\mathrm{dy}}{\mathrm{dx}}= \mathrm{slope} \mathrm{of} \mathrm{AB} \Rightarrow 2\mathrm{ax}+\mathrm{b}=\mathrm{a}(\mathrm{p}+\mathrm{q})+\mathrm{b} \\ \Rightarrow \mathrm{x}=\frac{\mathrm{p}+\mathrm{q}}{2} \end{gathered}$$