Problem
ALG-B1-M04-P022 Tangents to the Basic Parabola
The graph \(y=px^2+qx+r\) intersects the graph \(y=x^2\) at points with abscissas \(u\) and \(v\), \(u\ne v\). The tangents to \(y=x^2\) at these points meet at \(C\). If \(C\) lies on \(y=px^2+qx+r\), find \(p\).
Find the intersection of the tangents: \(\left(\frac{u+v}{2},uv\right)\).
The tangent to \(y=x^2\) at \(u\) is \(y=2ux-u^2\), and at \(v\) it is \(y=2vx-v^2\). Their intersection is \(x=\frac{u+v}{2}\), \(y=uv\).
Since the graphs intersect at \(u,v\), the polynomial \((p-1)x^2+qx+r\) has roots \(u,v\). Hence \((p-1)x^2+qx+r=(p-1)(x-u)(x-v)\).
Substitute \(C\) into \(y=px^2+qx+r\). Using \(q=-(p-1)(u+v)\), \(r=(p-1)uv\), we get \(uv=p\left(\frac{u+v}{2}\right)^2-(p-1)(u+v)\frac{u+v}{2}+(p-1)uv\). Simplification gives \((p-2)(u-v)^2=0\). Since \(u\ne v\), \(p=2\).
The graph problem is reduced to coefficients of a quadratic polynomial.