| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic translation and transformation |
| Difficulty | Standard +0.8 This FP1 question requires understanding parabola translations and then solving a discriminant problem. Part (a) is straightforward substitution (3 marks), but part (b) requires forming a quadratic in y by substituting x=ky into the translated parabola equation, then applying discriminant conditions for no real solutions—a multi-step problem requiring algebraic manipulation and careful reasoning about when curves don't intersect. The 6-mark allocation and discriminant analysis elevate this above routine exercises. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02w Graph transformations: simple transformations of f(x)1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(C: (y-3)^2 = 4a(x-2)\) | M1 | \((y-3)^2 = 4a(x-2)\) OE PI by next line |
| \((7-3)^2 = 4a(4-2)\) | A1 | OE |
| \(16 = 8a\), \(a = 2\) | A1 | AG Be convinced |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \((y-3)^2 = 4a(ky-2)\) | M1 | Replacing \(x\) by \(ky\) or \(y\) by \(\frac{x}{k}\) in \(c\)'s eqn for \(C\) |
| \(y^2 - (6+4ak)y + 9 + 8a = 0\) | A1 | A correct quadratic eqn in the form either \(Ay^2 + By + C = 0\) or \(Ax^2 + Bx + C = 0\) PI by later work |
| \(B^2 - 4AC = [-(6+4ak)]^2 - 4(9 + 8a)\) | M1 | \(B^2 - 4AC\) in terms of \(k\) (condone \(a\) remaining); ft on c's quadratic provided relevant coefficient(s) are in terms of \(k\) and \(A, B, C\) are all non zero |
| Roots non-real \(\Rightarrow B^2 - 4AC < 0 \Rightarrow [-(6+8k)]^2 - 4(25) < 0\) | A1 | A correct strict inequality where \(k\) is the only unknown (\(a\) must be replaced by 2 by this stage) |
| \((4k+8)(4k-2) < 0\); (*) critical values \(k = -2, k = 0.5\) | A1 | Correct critical values stated or used and correctly obtained |
| \(((*)\text{ true for}) -2 < k < 0.5\), (the values for which line does not meet curve \(C\).) | A1 | \(-2 < k < 0.5\) |
**(a)** $C: (y-3)^2 = 4a(x-2)$ | M1 | $(y-3)^2 = 4a(x-2)$ OE PI by next line
$(7-3)^2 = 4a(4-2)$ | A1 | OE
$16 = 8a$, $a = 2$ | A1 | AG Be convinced
**Total for (a): 3 marks**
**(b)** $(y-3)^2 = 4a(ky-2)$ | M1 | Replacing $x$ by $ky$ or $y$ by $\frac{x}{k}$ in $c$'s eqn for $C$
$y^2 - (6+4ak)y + 9 + 8a = 0$ | A1 | A correct quadratic eqn in the form either $Ay^2 + By + C = 0$ or $Ax^2 + Bx + C = 0$ PI by later work
$B^2 - 4AC = [-(6+4ak)]^2 - 4(9 + 8a)$ | M1 | $B^2 - 4AC$ in terms of $k$ (condone $a$ remaining); ft on c's quadratic provided relevant coefficient(s) are in terms of $k$ and $A, B, C$ are all non zero
Roots non-real $\Rightarrow B^2 - 4AC < 0 \Rightarrow [-(6+8k)]^2 - 4(25) < 0$ | A1 | A correct strict inequality where $k$ is the only unknown ($a$ must be replaced by 2 by this stage)
$(4k+8)(4k-2) < 0$; (*) critical values $k = -2, k = 0.5$ | A1 | Correct critical values stated or used and correctly obtained
$((*)\text{ true for}) -2 < k < 0.5$, (the values for which line does not meet curve $C$.) | A1 | $-2 < k < 0.5$
**Total for (b): 6 marks**
**Overall Total: 9 marks**
**Alternative (a):**
- (4,7) from translating $(4-2,7-3)$ ie $(2,4)$ on parabola $y^2 = 4ax$ (M1); $4^2 = 4a(2)$ (A1); $a=2(A1_{\text{above}})$
**Alternative (b):**
- Quadratic in $x$: eg $\frac{x^3}{k^2} - \left(\frac{6}{k}+4a\right)x + 9 + 8a = 0$
- Translating the line backwards to link with $y^2 = 4ax$: ie working with $y^2 = 4ax$ and $k(y+3) = x + 2$ gives eg $y^2 - 4aky - 12ak + 8a = 0$ then $2k^2 + 3k - 2 < 0$ etc
---A parabola with equation $y^2 = 4ax$, where $a$ is a constant, is translated by the vector $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ to give the curve $C$. The curve $C$ passes through the point $(4, 7)$.
\begin{enumerate}[label=(\alph*)]
\item Show that $a = 2$.
[3 marks]
\item Find the values of $k$ for which the line $ky = x$ does not meet the curve $C$.
[6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2016 Q6 [9]}}