| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola normal equation derivation |
| Difficulty | Challenging +1.2 This is a standard Further Pure 1 parabola question requiring parametric differentiation to find the normal equation, substitution back into the parabola equation, and algebraic manipulation. While it involves multiple steps and some algebraic complexity, the techniques are routine for FP1 students and the question provides significant scaffolding through its two-part structure. It's moderately harder than average A-level due to being Further Maths content, but represents a typical textbook exercise rather than requiring novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dy}{dx}=\frac{2a}{2ap}=\frac{1}{p}\) (or equivalent via \(y=2\sqrt{a}\sqrt{x}\) or \(2y\frac{dy}{dx}=4a\)) | B1 | Deduces correct tangent gradient |
| \(y-2ap=-p(x-ap^2)\) | M1 | Correct strategy for equation of normal (product of gradients \(=-1\)) |
| \(2aq-2ap=-p(aq^2-ap^2)\) leading to \(pq^2+2q-2p-p^3=0\) | A1 | Correct equation in terms of \(p\) and \(q\) |
| \((q-p)(pq+p^2+2)=0 \Rightarrow q=...\) | M1 | Applies correct strategy for finding \(q\); uses \(q=p\) as known root, inspection or long division |
| \(q=\frac{-p^2-2}{p}\) * | A1* | Correct proof with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(PQ^2=(ap^2-aq^2)^2+(2ap-2aq)^2\) | M1 | Sets up correct distance squared expression |
| \(=a^2(p-q)^2(p+q)^2+4a^2(p-q)^2\) | M1 | Correct algebraic manipulation |
| \(=a^2(p-q)^2\left[(p+q)^2+4\right]\) | A1 | Correct factored form |
| \(=a^2\left(2p+\frac{2}{p}\right)^2\left[\left(-\frac{2}{p}\right)^2+4\right]\) | Substitutes \(q=\frac{-p^2-2}{p}\) | |
| \(=\frac{4a^2}{p^2}(p^2+1)^2\cdot\frac{4}{p^2}(p^2+1)=\frac{16a^2}{p^4}(p^2+1)^3\) | A1, A1 | Each correct step to final result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Applies Pythagoras correctly to find \(PQ^2\) | M1 | |
| Uses their \(q\) in terms of \(p\) to obtain an expression in terms of \(p\) only | M1 | |
| Correct expression in any form in terms of \(p\) only | A1 | |
| \(k = 16\) or \(n = 3\) | A1 | |
| \(k = 16\) and \(n = 3\) | A1 |
## Question 5:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx}=\frac{2a}{2ap}=\frac{1}{p}$ (or equivalent via $y=2\sqrt{a}\sqrt{x}$ or $2y\frac{dy}{dx}=4a$) | B1 | Deduces correct tangent gradient |
| $y-2ap=-p(x-ap^2)$ | M1 | Correct strategy for equation of normal (product of gradients $=-1$) |
| $2aq-2ap=-p(aq^2-ap^2)$ leading to $pq^2+2q-2p-p^3=0$ | A1 | Correct equation in terms of $p$ and $q$ |
| $(q-p)(pq+p^2+2)=0 \Rightarrow q=...$ | M1 | Applies correct strategy for finding $q$; uses $q=p$ as known root, inspection or long division |
| $q=\frac{-p^2-2}{p}$ * | A1* | Correct proof with no errors |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $PQ^2=(ap^2-aq^2)^2+(2ap-2aq)^2$ | M1 | Sets up correct distance squared expression |
| $=a^2(p-q)^2(p+q)^2+4a^2(p-q)^2$ | M1 | Correct algebraic manipulation |
| $=a^2(p-q)^2\left[(p+q)^2+4\right]$ | A1 | Correct factored form |
| $=a^2\left(2p+\frac{2}{p}\right)^2\left[\left(-\frac{2}{p}\right)^2+4\right]$ | | Substitutes $q=\frac{-p^2-2}{p}$ |
| $=\frac{4a^2}{p^2}(p^2+1)^2\cdot\frac{4}{p^2}(p^2+1)=\frac{16a^2}{p^4}(p^2+1)^3$ | A1, A1 | Each correct step to final result |
## Question (b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Applies Pythagoras correctly to find $PQ^2$ | M1 | |
| Uses their $q$ in terms of $p$ to obtain an expression in terms of $p$ only | M1 | |
| Correct expression in any form in terms of $p$ only | A1 | |
| $k = 16$ or $n = 3$ | A1 | |
| $k = 16$ and $n = 3$ | A1 | |\begin{enumerate}
\item The point $P \left( a p ^ { 2 } , 2 a p \right)$, where $a$ is a positive constant, lies on the parabola with equation
\end{enumerate}
$$y ^ { 2 } = 4 a x$$
The normal to the parabola at $P$ meets the parabola again at the point $Q \left( a q ^ { 2 } , 2 a q \right)$\\
(a) Show that
$$q = \frac { - p ^ { 2 } - 2 } { p }$$
(b) Hence show that
$$P Q ^ { 2 } = \frac { k a ^ { 2 } } { p ^ { 4 } } \left( p ^ { 2 } + 1 \right) ^ { n }$$
where $k$ and $n$ are integers to be determined.
\hfill \mbox{\textit{Edexcel FP1 AS 2021 Q5 [10]}}