A2 June 2022 Paper 2 Q10
10 The curve \(C_1\) has equation
\[\frac{x^2}{25} - \frac{y^2}{4} = 1\]The curve \(C_2\) has equation
\[x^2 - 25y^2 - 6x - 200y - 416 = 0\]Give your answers in the form \(ax + by + c = 0\) where \(a\), \(b\) and \(c\) are integers. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Deduces that a stretch, only along the \(y\)-axis, is required | M1 | 2.2a |
| Deduces that a translation with two non-zero components is needed (or a translation parallel to the \(x\)-axis and a translation parallel to the \(y\)-axis) | M1 | 2.2a |
| Obtains correct stretch and/or correct translation (accept \(\begin{bmatrix}3 \\ -8\end{bmatrix}\)) | A1 | 1.1b |
| States correct order of completely correct transformations (accept any correct sequence) | E1 | 2.4 |
| (4) |
Typical solution
\[\frac{x^2}{25} - \frac{y^2}{4} = 1\]Stretch s.f. ½ along \(y\)-axis gives
\[\frac{x^2}{25} - \frac{(2y)^2}{4} = 1\]\[\frac{x^2}{25} - y^2 = 1\]Translation by vector \(\begin{bmatrix}3 \\ -4\end{bmatrix}\) gives
\[\frac{(x - 3)^2}{25} - (y + 4)^2 = 1\]which is equivalent to the equation for \(C_2\)
Stretch, scale factor ½, along \(y\)-axis followed by translation by vector \(\begin{bmatrix}3 \\ -4\end{bmatrix}\)
| Scheme | Marks | AO |
|---|---|---|
| Obtains the correct asymptotes of \(C_1\) (PI) or uses a correct method to obtain both asymptotes of \(C_2\) directly PI by an equation of the form \(5y \pm x + k = 0\) | M1 | 1.1a |
| Correctly applies their sequence of at least two different types of transformations to one asymptote or obtains an equation of the form \(5y \pm x + k = 0\) where \(k\) is a non-zero integer constant | M1 | 3.1a |
| Obtains both correct results | A1 | 1.1b |
| (3) | ||
| (7 marks) |
Typical solution
Asymptotes of \(C_1\): \(y = \pm\dfrac{2x}{5}\)
\[y = \frac{2x}{5}\]Stretch \(\Rightarrow 2y = \dfrac{2x}{5} \Rightarrow 5y = x\)
Translation \(\Rightarrow 5(y + 4) = x - 3\)
\[5y - x + 23 = 0\]\[y = -\frac{2x}{5}\]Stretch \(\Rightarrow 2y = -\dfrac{2x}{5} \Rightarrow 5y = -x\)
Translation \(\Rightarrow 5(y + 4) = -x - 3\)
\[5y + x + 17 = 0\]