FP1 June 2014 (R) Q8
8. The rectangular hyperbola \(H\) has equation \(xy = c^2\), where \(c\) is a positive constant.
The point \(P\left(ct, \dfrac{c}{t}\right)\), \(t \neq 0\), is a general point on \(H\).
An equation for the tangent to \(H\) at \(P\) is given by \[y = -\frac{1}{t^2}x + \frac{2c}{t}\]
The points \(A\) and \(B\) lie on \(H\).
The tangent to \(H\) at \(A\) and the tangent to \(H\) at \(B\) meet at the point \(\left(-\dfrac{6}{7}c, \dfrac{12}{7}c\right)\).
Find, in terms of \(c\), the coordinates of \(A\) and the coordinates of \(B\). (5)
| Scheme | Marks |
|---|---|
| \(\dfrac{12}{7}c = -\dfrac{1}{t^2} \times -\dfrac{6}{7}c + \dfrac{2c}{t}\) Substitutes \(\left(-\dfrac{6}{7}c, \dfrac{12}{7}c\right)\) into the equation of the tangent | M1 |
| \(\dfrac{12}{7}c = -\dfrac{1}{t^2} \times -\dfrac{6}{7}c + \dfrac{2c}{t} \Rightarrow\) \(6t^2 - 7t - 3 = 0\) Correct 3TQ in terms of \(t\) | A1 |
| \(6t^2 - 7t - 3 = 0 \Rightarrow (3t + 1)(2t - 3) = 0 \Rightarrow t =\) Attempt to solve their 3TQ for \(t\) | M1 |
| \(t = -\dfrac{1}{3}, t = \dfrac{3}{2} \Rightarrow \left(-\dfrac{1}{3}c, -3c\right), \left(\dfrac{3}{2}c, \dfrac{2}{3}c\right)\) M1: Uses at least one of their values of \(t\) to find \(A\) or \(B\). A1: Correct coordinates. | M1A1 |
| (5) | |
| (5 marks) |