June 2022 Paper 1 Q12
12 A curve has parametric equations \(x = \dfrac{1}{t}\), \(y = 2t\). The point \(P\) is \(\left(\dfrac{1}{p}, 2p\right)\).
The tangent to this curve at \(P\) crosses the \(x\)-axis at the point \(A\) and the normal to this curve at \(P\) crosses the \(x\)-axis at the point \(B\).
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{-1}{t^2}\), \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 2\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\;\frac{\mathrm{d}y}{\mathrm{d}t}\;}{\;\frac{\mathrm{d}x}{\mathrm{d}t}\;}\) | M1 | 1.1a |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -2t^2\) | A1 | 2.1 |
| \(y - 2p = -2p^2\left(x - \tfrac{1}{p}\right)\) | M1 | 1.1a |
| \(y = -2p^2x + 4p\) A.G. | A1 | 2.1 |
| [4] |
Notes
M1: Attempt correct process to find gradient in terms of \(t\) or \(p\)
Correctly combine attempts at two derivatives
Need \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = kt^{-2}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 2\)
SC B1 for gradient of \(-2x^{-2}\) if it is never seen in terms of \(t\) or \(p\)
A1: Obtain correct gradient
In terms of \(t\) or \(p\)
M1: Attempt equation of tangent
Condone still working in terms of \(t\)
Allow mixture of \(t\) and \(p\) as long as convincingly recovered
Using their gradient from a differentiation attempt, but not dependent on first M1
Substitution into \(y - y_1 = m(x - x_1)\) or equation involving \(c\) from \(y = mx + c\)
A1: Obtain given answer
Must now be in terms of \(p\)
Expand brackets and simplify to given answer, or find \(c\) and substitute back into equation
| Scheme | Marks | AO |
|---|---|---|
| \(m' = \frac{1}{2p^2}\) | B1FT | 1.1a |
| \(y - 2p = \frac{1}{2p^2}\left(x - \frac{1}{p}\right)\) \(y = \frac{1}{2p^2}x + 2p - \frac{1}{2p^3}\) | M1 | 1.1 |
| Use \(y = 0\) to attempt \(x\)-coordinate of \(B\) | M1 | 3.1a |
| at \(B\), \(y = 0\) so \(x = 2p^2\left(\frac{1}{2p^3} - 2p\right) = \frac{1}{p} - 4p^3\) | A1 | 2.1 |
| at \(A\), \(y = 0\) so \(x = \frac{4p}{2p^2} = \frac{2}{p}\) | B1 | 2.1 |
| \(PA = \sqrt{\left(\frac{1}{p}\right)^2 + (2p)^2}\) \(PB = \sqrt{\left(4p^3\right)^2 + (2p)^2}\) | M1 | 3.1a |
| Correct \(PA\) and \(PB\) | A1 | 2.1 |
| \(PA : PB = \frac{1}{p}\sqrt{4p^4 + 1} : 2p\sqrt{4p^4 + 1}\) \(= \frac{1}{p} : 2p\) \(= 1 : 2p^2\) A.G. | A1 | 2.1 |
| [8] |
Notes
B1FT: Correct (unsimplified) gradient of normal, following their derivative
Gradient in terms of \(t\) or \(p\), but not \(x\)
Could either FT on their incorrect derivative or deduce the gradient from the equation given in (a)
M1: Attempt equation of normal
Attempt to use their gradient and \(P\)
Allow mixture of \(t\) and \(p\) as long as convincingly recovered
Substitution into \(y - y_1 = m(x - x_1)\) or equation involving \(c\) from \(y = mx + c\)
M1: Using their attempt at normal equation
As far as finding an expression for \(x\)
A1: Correct \(x\)-coordinate for \(B\)
Any equivalent form
B1: Correct \(x\)-coordinate for \(A\)
Any equivalent form
M1: Attempt length of \(PA\) or \(PB\)
Or M1 for attempting one of \((PA)^2\) or \((PB)^2\)
Must correct distance formula
Using the given \(P\), and their coordinates for \(A\) and/or \(B\), which must involve a function of \(p\)
A1: Or correct \((PA)^2\) and \((PB)^2\)
A1: Simplify ratio to obtain given answer
Must show clear method, such as same expression in each square root before cancelling
Could also consider fraction and then cancel to deduce given ratio
Could simplify \((PA)^2 : (PB)^2\), and then square root to obtain ratio