Problem
ALG-B1-M04-P019 Width of a Parabola
A parabola passes through \((a,0)\), \((b,0)\), where \(a
The vertex lies midway between the roots and has ordinate \(-1\).
The axis of the parabola passes through the midpoint of the roots, \(c=\frac{a+b}{2}\), and the vertex has coordinates \((c,-1)\). Hence \(y=\lambda(x-c)^2-1\). Substituting \(x=a\) gives \(0=\lambda\left(\frac{b-a}{2}\right)^2-1\), so \(\lambda=\frac{4}{(b-a)^2}\).
For two parabolas, common points satisfy \(\frac{4}{w_1^2}(x-c_1)^2=\frac{4}{w_2^2}(x-c_2)^2\). This is equivalent to two linear equations \(\frac{x-c_1}{w_1}=\pm\frac{x-c_2}{w_2}\). If \(w_1\ne w_2\) and \(c_1\ne c_2\), both give one distinct solution. If the width or the axis coincides, one solution merges or disappears, leaving exactly one common point.
This is a local version of the graph problem: the main point is algebraic parametrisation of the parabola.