FP1 June 2016 Q9
9. The rectangular hyperbola, \(H\), has cartesian equation \(xy = 25\)
This normal meets the line with equation \(y = -x\) at the point \(A\).
The point \(M\) is the midpoint of the line segment \(AP\).
Given that \(M\) lies on the positive \(x\)-axis,
| Scheme | Marks |
|---|---|
| \(y = \dfrac{25}{x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -25x^{-2}\), or \(y + x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{y}{x}\) or \(\dot{x} = 5, \dot{y} = -\dfrac{5}{t^2}\) so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{t^2}\) | B1 |
| and at \(P\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{p^2}\) so gradient of normal is \(p^2\) | M1 A1 |
| Either \(y - \dfrac{5}{p} = p^2(x - 5p)\) or \(y = p^2x + k\) and use \(x = 5p\), \(y = \dfrac{5}{p}\) | M1 |
| \(\Rightarrow y - p^2x = \dfrac{5}{p} - 5p^3\) (*) | A1cso |
| (5) |
Notes
B1: Any correct expression for gradient of tangent
M1: Substitutes values into derived expression using calculus to give gradient of normal at \(P\)
A1: cao. Can be implied by use in equation of a straight line
M1: Use of formula for the equation of a straight line with their changed gradient
A1: cso
| Scheme | Marks |
|---|---|
| At the point \(A\): \(y + p^2y = \dfrac{5}{p} - 5p^3\) or \(-x - p^2x = \dfrac{5}{p} - 5p^3\) | |
| \(y(1 + p^2) = \dfrac{5}{p}(1 - p^4)\) or \(-x(1 + p^2) = \dfrac{5}{p}(1 - p^4)\) | M1 |
| \(y = \dfrac{\frac{5}{p}(1 - p^2)(1 + p^2)}{(1 + p^2)} = \dfrac{5}{p}(1 - p^2) = \dfrac{5}{p} - 5p\) or \(x = \dfrac{-\frac{5}{p}(1 - p^2)(1 + p^2)}{(1 + p^2)} = \dfrac{-5}{p}(1 - p^2) = \dfrac{-5}{p} + 5p\) * | M1 |
| so \(x = -\dfrac{5}{p}(1 - p^2) = -\dfrac{5}{p} + 5p\) and \(y = \dfrac{5}{p} - 5p\) * | A1cso |
| (3) |
Notes
M1: Replaces \(x\) by \(-y\) or \(y\) by \(-x\)
M1: Factorises \((1 - p^4)\) to simplify answer in first variable
A1 cso: Obtains both \(x\) and \(y\)
ALT (b) Accept Verification.
M1: Substitutes the coordinates of \(A\) into the equation of the normal
M1: Substitutes the coordinates of \(A\) into both the normal and \(y = -x\).
A1 cso: No errors seen
| Scheme | Marks |
|---|---|
| \(M\) has coordinates \(\left(-\dfrac{5}{2p} + 5p, \dfrac{5}{p} - \dfrac{5p}{2}\right)\) o.e. | B1 |
| So when \(y = 0\), \(\dfrac{5}{p} - \dfrac{5p}{2} = 0\) and \(p = \sqrt{2}\) so \(M\) has \(x\) coordinate \(\dfrac{15}{4}\sqrt{2}\) o.e. | M1 A1 |
| (3) | |
| (11 marks) |
Notes
B1: Correct \(x\)- coordinate of midpoint (may be implied) and correct \(y\) coordinate, accept equivalent forms
M1: Puts their \(y = 0\) and finds value for \(p\) to use in \(x =\)
A1: \(+\dfrac{15}{4}\sqrt{2}\) or equivalent only