June 2024 Paper 3 Q18

18 This question refers to the article on the Insert, “Tangents and normals to a quadratic curve”. The relevant extract (lines 11 to 20) is reproduced here.

The general quadratic curve has equation \(y = ax^2 + bx + c\). The tangents at any two points P and Q on this curve also cross at a point whose \(x\)-coordinate is equal to the mean of the \(x\)-coordinates of P and Q. So if P has \(x\)-coordinate \(x_\mathrm{P}\) and Q has \(x\)-coordinate \(x_\mathrm{Q}\) then the \(x\)-coordinate of the intersection point of the tangents is \(\dfrac{x_\mathrm{P} + x_\mathrm{Q}}{2}\). The \(y\)-coordinate of the intersection point can be shown to be \(ax_\mathrm{P}x_\mathrm{Q} + b\left(\frac{x_\mathrm{P}+x_\mathrm{Q}}{2}\right) + c\). This is equivalent to \(a\left(\frac{x_\mathrm{P}+x_\mathrm{Q}}{2}\right)^2 + b\left(\frac{x_\mathrm{P}+x_\mathrm{Q}}{2}\right) + c - a\left(\frac{x_\mathrm{P}-x_\mathrm{Q}}{2}\right)^2\). The formula \(ax_\mathrm{P}x_\mathrm{Q} + b\left(\frac{x_\mathrm{P}+x_\mathrm{Q}}{2}\right) + c\) looks simpler but \(y = a\left(\frac{x_\mathrm{P}+x_\mathrm{Q}}{2}\right)^2 + b\left(\frac{x_\mathrm{P}+x_\mathrm{Q}}{2}\right) + c - a\left(\frac{x_\mathrm{P}-x_\mathrm{Q}}{2}\right)^2\) is in terms of the \(x\)-coordinate of the point of intersection, apart from the last term. For pairs of points with \(x_\mathrm{P} - x_\mathrm{Q} = h\) where \(h\) is a constant, the point of intersection of the tangents lies on the curve \(y = ax^2 + bx + c - \dfrac{ah^2}{4}\). This curve is a translation of the original curve \(y = ax^2 + bx + c\).

A student is investigating the intersection points of tangents to the curve \(y = 6x^2 - 7x + 1\). She uses software to draw tangents at pairs of points with \(x\)-coordinates differing by 5.

Find the equation of the curve that all the intersection points lie on. [2]