June 2024 Paper 3 Q13

13 This question refers to the article on the Insert, “Tangents and normals to a quadratic curve”. The relevant extract (lines 1 to 10) is reproduced here.

Tangents

Fig. C1 shows the curve \(y = x^2\) together with tangents to the curve at points A \((-3, 9)\) and B \((1, 1)\). The tangents cross at the point \((-1, -3)\). This has \(x\)-coordinate \(-1\), which is the mean of the \(x\)-coordinates of points A and B.

For the curve \(y = x^2\), the equation of the tangent at a general point \((t, t^2)\) is \(y = 2tx - t^2\). So the equation of the tangent at the point \((t_1, t_1^2)\) is \(y = 2t_1x - t_1^2\). There is a similar equation for the tangent at the point \((t_2, t_2^2)\), and these two tangents cross where \(2t_1x - t_1^2 = 2t_2x - t_2^2\).

This gives \(2x(t_1 - t_2) = t_1^2 - t_2^2\) so \(2x(t_1 - t_2) = (t_1 - t_2)(t_1 + t_2)\) hence \(x = \dfrac{t_1 + t_2}{2}\). The \(y\)-coordinate of the point of intersection is \(t_1t_2\).

Substitute appropriate values of \(t_1\) and \(t_2\) to verify that \(t_1t_2\) gives the correct value for the \(y\)-coordinate of the point of intersection of the tangents at the points A and B in Fig. C1. [1]