| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Rectangular hyperbola tangent intersection |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question on parabola tangents requiring parametric differentiation, simultaneous equations, and knowledge of directrix properties. While the techniques are standard for FP1, the four-part structure with algebraic manipulation and the final conceptual link to the directrix elevates it above typical A-level questions but remains within expected FP1 scope. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 2a^2x^{\frac{1}{2}} \Rightarrow \dfrac{dy}{dx} = a^2 x^{-\frac{1}{2}}\) | M1 | \(ky\dfrac{dy}{dx} = c\); or \(\dfrac{dy}{dt}\times\dfrac{1}{\frac{dx}{dt}}\), can be function of \(p\) or \(t\) |
| \(y^2 = 4ax \Rightarrow 2y\dfrac{dy}{dx} = 4a\) | ||
| \(\dfrac{dy}{dx} = a^2 x^{-\frac{1}{2}}\) or \(2y\dfrac{dy}{dx}=4a\) or \(\dfrac{dy}{dx}=2a\cdot\dfrac{1}{2ap}\) | A1 | Differentiation is accurate |
| \(y - 2ap = \dfrac{1}{p}(x - ap^2)\) | M1 | Applies \(y-2ap =\) their \(m(x-ap^2)\); m must be a function of p from calculus |
| \(py - x = ap^2\) * | A1 cso | Correct completion to printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(qy - x = aq^2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(qy - aq^2 = py - ap^2\) | M1 | Attempt to obtain an equation in one variable \(x\) or \(y\) |
| \(y(q-p) = aq^2 - ap^2\) | M1 | Attempt to isolate \(x\) or \(y\) |
| \(y = a(p+q)\) or \(ap+aq\) | A1 | Either one correct simplified coordinate |
| \(x = apq\) | A1 | Both correct simplified coordinates |
| \(R(apq,\ ap+aq)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(apq = -a\) | M1 | Their \(x\) coordinate of \(R = -a\) |
| \(pq = -1\) | A1 | Answer only: scores 2/2 if \(x\) coordinate of \(R\) is \(apq\), otherwise 0/2 |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 2a^2x^{\frac{1}{2}} \Rightarrow \dfrac{dy}{dx} = a^2 x^{-\frac{1}{2}}$ | M1 | $ky\dfrac{dy}{dx} = c$; or $\dfrac{dy}{dt}\times\dfrac{1}{\frac{dx}{dt}}$, can be function of $p$ or $t$ |
| $y^2 = 4ax \Rightarrow 2y\dfrac{dy}{dx} = 4a$ | | |
| $\dfrac{dy}{dx} = a^2 x^{-\frac{1}{2}}$ or $2y\dfrac{dy}{dx}=4a$ or $\dfrac{dy}{dx}=2a\cdot\dfrac{1}{2ap}$ | A1 | Differentiation is accurate |
| $y - 2ap = \dfrac{1}{p}(x - ap^2)$ | M1 | Applies $y-2ap =$ their $m(x-ap^2)$; **m must be a function of p from calculus** |
| $py - x = ap^2$ * | A1 cso | Correct completion to printed answer |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $qy - x = aq^2$ | B1 | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $qy - aq^2 = py - ap^2$ | M1 | Attempt to obtain an equation in one variable $x$ or $y$ |
| $y(q-p) = aq^2 - ap^2$ | M1 | Attempt to isolate $x$ or $y$ |
| $y = a(p+q)$ or $ap+aq$ | A1 | Either one correct simplified coordinate |
| $x = apq$ | A1 | Both correct simplified coordinates |
| $R(apq,\ ap+aq)$ | | |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $apq = -a$ | M1 | Their $x$ coordinate of $R = -a$ |
| $pq = -1$ | A1 | **Answer only**: scores 2/2 if $x$ coordinate of $R$ is $apq$, otherwise 0/2 |6. A parabola $C$ has equation $y ^ { 2 } = 4 a x , \quad a > 0$
The points $P \left( a p ^ { 2 } , 2 a p \right)$ and $Q \left( a q ^ { 2 } , 2 a q \right)$ lie on $C$, where $p \neq 0 , q \neq 0 , p \neq q$.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation of the tangent to the parabola at $P$ is
$$p y - x = a p ^ { 2 }$$
\item Write down the equation of the tangent at $Q$.
The tangent at $P$ meets the tangent at $Q$ at the point $R$.
\item Find, in terms of $p$ and $q$, the coordinates of $R$, giving your answers in their simplest form.
Given that $R$ lies on the directrix of $C$,
\item find the value of $p q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q6 [11]}}