October 2021 Paper 1 Q13
13. A curve \(C\) has parametric equations
\[x = \frac{t^2 + 5}{t^2 + 1} \qquad y = \frac{4t}{t^2 + 1} \qquad t \in \mathbb{R}\]Show that all points on \(C\) satisfy\[(x - 3)^2 + y^2 = 4\](3)
| Scheme | Marks | AO |
|---|---|---|
| \((x - 3)^2 + y^2 = \left(\dfrac{t^2 + 5}{t^2 + 1} - 3\right)^2 + \left(\dfrac{4t}{t^2 + 1}\right)^2\) | M1 | 3.1a |
| \(= \dfrac{\left(2 - 2t^2\right)^2 + 16t^2}{\left(t^2 + 1\right)^2} = \dfrac{4 + 8t^2 + 4t^4}{\left(t^2 + 1\right)^2}\) | dM1 | 1.1b |
| \(\dfrac{4\left(t^4 + 2t^2 + 1\right)}{\left(t^2 + 1\right)^2} = \dfrac{4\left(t^2 + 1\right)^2}{\left(t^2 + 1\right)^2} = 4\,*\) | A1* | 2.1 |
| (3) | ||
| (3 marks) |
Notes
M1: Attempts to substitute the given parametric forms into the Cartesian equation or the lhs of the Cartesian equation. There may have been an (incorrect) attempt to multiply out the \((x - 3)^2\) term.
dM1: Attempts to combine (at least the lhs) using correct processing into a single fraction, multiplies out and collects terms on the numerator.
A1*: Fully correct proof showing all key steps
Alt
| Scheme | Marks | AO |
|---|---|---|
| \(x = \dfrac{t^2 + 5}{t^2 + 1} \Rightarrow xt^2 + x = t^2 + 5 \Rightarrow t^2 = \dfrac{5 - x}{x - 1}\) \(y = \dfrac{4t}{t^2 + 1} \Rightarrow y^2 = \dfrac{16t^2}{\left(t^2 + 1\right)^2} = \dfrac{16\left(\dfrac{5 - x}{x - 1}\right)}{\left(\dfrac{5 - x}{x - 1} + 1\right)^2}\) | M1 | 3.1a |
| \(y^2 = \dfrac{16\left(\dfrac{5 - x}{x - 1}\right)}{\left(\dfrac{5 - x}{x - 1} + 1\right)^2} = 16\left(\dfrac{5 - x}{x - 1}\right) \times \left(\dfrac{(x - 1)}{5 - x + x - 1}\right)^2 \Rightarrow y^2 = (5 - x)(x - 1)\) | dM1 | 1.1b |
| \(y^2 = (5 - x)(x - 1) \Rightarrow y^2 = 6x - x^2 - 5\) \(\Rightarrow y^2 = 4 - (x - 3)^2\) or other intermediate step \(\Rightarrow (x - 3)^2 + y^2 = 4\,*\) | A1* | 2.1 |
| (3) |
M1: Adopts a correct strategy for eliminating \(t\) to obtain an equation in terms of \(x\) and \(y\) only. See scheme.
Other methods exist which also lead to an appropriate equation. E.g using \(t = \dfrac{y}{x - 1}\)
dM1: Uses correct processing to eliminate the fractions and start to simplify
A1*: Fully correct proof showing all key steps