| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola area calculations |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring parametric representation of parabolas, tangent equations using the given derivative formula, finding intersection points with the directrix, and calculating a specific area using integration. While the derivative is given and the steps are guided, it requires solid understanding of parabola properties, careful algebraic manipulation, and integration of a parabola—more demanding than standard C1-C4 content but routine for FP1 students. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y^2 = (8p)^2 = 64p^2\) and \(16x = 16(4p^2) = 64p^2\), \(\Rightarrow P(4p^2, 8p)\) is a general point on \(C\) | B1 | Substitutes \(y_p = 8p\) into \(y^2\) to obtain \(64p^2\) and substitutes \(x_p = 4p^2\) into \(16x\) to obtain \(64p^2\) and concludes that \(P\) lies on \(C\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y^2 = 16x\) gives \(a = 4\), or \(2y\frac{dy}{dx} = 16\) so \(\frac{dy}{dx} = \frac{8}{y}\) | M1 | Uses the given formula to deduce the derivative, or differentiates using chain rule |
| \(l: y - 8p = \left(\frac{8}{8p}\right)\left(x - 4p^2\right)\) | M1 | Applies \(y - 8p = m(x - 4p^2)\) with tangent gradient \(m\) in terms of \(p\). Accept use of \(8p = m(4p^2) + c\) with clear attempt to find \(c\) |
| leading to \(py = x + 4p^2\) * | A1* | Obtains \(py = x + 4p^2\) by cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(B\!\left(-4, \frac{10}{3}\right)\) into \(l \Rightarrow \frac{10p}{3} = -4 + 4p^2\) | M1 | Substitutes \(x = -a\) and \(y = \frac{10}{3}\) into \(l\) |
| \(6p^2 - 5p - 6 = 0 \Rightarrow (2p-3)(3p+2) = 0 \Rightarrow p = \ldots\) | M1 | Obtains a 3-term quadratic and solves to give \(p = \ldots\) |
| \(p = \frac{3}{2}\) and \(l\) cuts \(x\)-axis when \(\frac{3}{2}(0) = x + 4\!\left(\frac{3}{2}\right)^2 \Rightarrow x = \ldots\) | M1 | Substitutes their \(p\) (must be positive) and \(y = 0\) into \(l\) and solves for \(x\) |
| \(x = -9\) | A1 | Finds that \(l\) cuts the \(x\)-axis at \(x = -9\) |
| \(p = \frac{3}{2} \Rightarrow P(9,12) \Rightarrow \text{Area}(R) = \frac{1}{2}(9 - -9)(12) - \int_0^9 4x^{\frac{1}{2}}\,dx\) | M1 | Fully correct method for finding area of \(R\): \(\frac{1}{2}(\text{their } x_p - \text{``}-9\text{''})\!(\text{their } y_p) - \int_0^{\text{their } x_p} 4x^{\frac{1}{2}}\,dx\) |
| \(\int 4x^{\frac{1}{2}}\,dx = \frac{4x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}(+c)\) or \(\frac{8}{3}x^{\frac{3}{2}}(+c)\) | M1 | Integrates \(\pm\lambda x^{\frac{1}{2}}\) to give \(\pm\mu x^{\frac{3}{2}}\), where \(\lambda, \mu \neq 0\) |
| \(\frac{8}{3}x^{\frac{3}{2}}(+c)\) | A1 | Integrates \(4x^{\frac{1}{2}}\) to give \(\frac{8}{3}x^{\frac{3}{2}}\), simplified or unsimplified |
| \(\text{Area}(R) = \frac{1}{2}(18)(12) - \frac{8}{3}\!\left(9^{\frac{3}{2}} - 0\right) = 108 - 72 = 36\) * | A1* | Fully correct proof leading to correct answer of 36 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(B\!\left(-4,\frac{10}{3}\right)\) into \(l \Rightarrow \frac{10p}{3} = -4 + 4p^2\) | M1 | Substitutes \(x = -a\) and \(y = \frac{10}{3}\) into \(l\) |
| \(6p^2 - 5p - 6 = 0 \Rightarrow (2p-3)(3p+2) = 0 \Rightarrow p = \ldots\) | M1 | Obtains 3-term quadratic and solves |
| \(p = \frac{3}{2}\) into \(l\) gives \(\frac{3}{2}y = x + 4\!\left(\frac{3}{2}\right)^2 \Rightarrow x = \ldots\) | M1 | Substitutes their \(p\) into \(l\) and rearranges to give \(x = \ldots\) |
| \(x = \frac{3}{2}y - 9\) | A1 | |
| \(p = \frac{3}{2} \Rightarrow P(9,12) \Rightarrow \text{Area}(R) = \int_0^{12}\!\left(\frac{1}{16}y^2 - \left(\frac{3}{2}y - 9\right)\right)dy\) | M1 | Fully correct method for area of \(R\): \(\int_0^{\text{their } y_p}\!\left(\frac{1}{16}y^2 - \text{their}\!\left(\frac{3}{2}y - 9\right)\right)dy\) |
| \(\int\!\left(\frac{1}{16}y^2 - \frac{3}{2}y + 9\right)dy = \frac{1}{48}y^3 - \frac{3}{4}y^2 + 9y\,(+c)\) | M1, A1 | Integrates \(\pm\lambda y^2 \pm \mu y \pm \nu\) to give \(\pm\alpha y^3 \pm \beta y^2 \pm \nu y\); correct integration |
| \(\text{Area}(R) = \left(\frac{1}{48}(12)^3 - \frac{3}{4}(12)^2 + 9(12)\right) - (0) = 36 - 108 + 108 = 36\) * | A1* | Fully correct proof leading to 36 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(B\!\left(-4,\frac{10}{3}\right)\) into \(l \Rightarrow \frac{10p}{3} = -4 + 4p^2\) | M1 | Substitutes \(x = -a\) and \(y = \frac{10}{3}\) into \(l\) |
| \(6p^2 - 5p - 6 = 0 \Rightarrow (2p-3)(3p+2) = 0 \Rightarrow p = \ldots\) | M1 | Obtains 3-term quadratic and solves |
| \(p = \frac{3}{2}\) and \(l\) cuts \(x\)-axis when \(\frac{3}{2}(0) = x + 4\!\left(\frac{3}{2}\right)^2 \Rightarrow x = \ldots\) | M1 | Substitutes their \(p\) (must be positive) and \(y=0\) into \(l\) |
| \(x = -9\) | A1 | |
| \(p = \frac{3}{2} \Rightarrow P(9,12)\) and \(x=0\) in \(l\): \(y = \frac{2}{3}x + 6\) gives \(y = 6\); \(\Rightarrow \text{Area}(R) = \frac{1}{2}(9)(6) + \int_0^9\!\left(\left(\frac{2}{3}x + 6\right) - \left(4x^{\frac{1}{2}}\right)\right)dx\) | M1 | Fully correct method for area of \(R\) |
| \(\int\!\left(\frac{2}{3}x + 6 - 4x^{\frac{1}{2}}\right)dx = \frac{1}{3}x^2 + 6x - \frac{8}{3}x^{\frac{3}{2}}\,(+c)\) | M1, A1 | Integrates \(\pm\lambda x \pm \mu \pm \nu x^{\frac{1}{2}}\) correctly |
| \(\text{Area}(R) = 27 + \left(\left(\frac{1}{3}(9)^2 + 6(9) - \frac{8}{3}(9^{\frac{3}{2}})\right) - (0)\right) = 27 + (27 + 54 - 72) = 27 + 9 = 36\) * | A1* | Fully correct proof leading to 36 |
# Question 5:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y^2 = (8p)^2 = 64p^2$ and $16x = 16(4p^2) = 64p^2$, $\Rightarrow P(4p^2, 8p)$ is a general point on $C$ | B1 | Substitutes $y_p = 8p$ into $y^2$ to obtain $64p^2$ and substitutes $x_p = 4p^2$ into $16x$ to obtain $64p^2$ and concludes that $P$ lies on $C$ |
**(1 mark)**
---
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y^2 = 16x$ gives $a = 4$, or $2y\frac{dy}{dx} = 16$ so $\frac{dy}{dx} = \frac{8}{y}$ | M1 | Uses the given formula to deduce the derivative, or differentiates using chain rule |
| $l: y - 8p = \left(\frac{8}{8p}\right)\left(x - 4p^2\right)$ | M1 | Applies $y - 8p = m(x - 4p^2)$ with tangent gradient $m$ in terms of $p$. Accept use of $8p = m(4p^2) + c$ with clear attempt to find $c$ |
| leading to $py = x + 4p^2$ * | A1* | Obtains $py = x + 4p^2$ by **cso** |
**(3 marks)**
---
## Part (c) — Main Method:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $B\!\left(-4, \frac{10}{3}\right)$ into $l \Rightarrow \frac{10p}{3} = -4 + 4p^2$ | M1 | Substitutes $x = -a$ and $y = \frac{10}{3}$ into $l$ |
| $6p^2 - 5p - 6 = 0 \Rightarrow (2p-3)(3p+2) = 0 \Rightarrow p = \ldots$ | M1 | Obtains a 3-term quadratic and solves to give $p = \ldots$ |
| $p = \frac{3}{2}$ and $l$ cuts $x$-axis when $\frac{3}{2}(0) = x + 4\!\left(\frac{3}{2}\right)^2 \Rightarrow x = \ldots$ | M1 | Substitutes their $p$ (must be positive) and $y = 0$ into $l$ and solves for $x$ |
| $x = -9$ | A1 | Finds that $l$ cuts the $x$-axis at $x = -9$ |
| $p = \frac{3}{2} \Rightarrow P(9,12) \Rightarrow \text{Area}(R) = \frac{1}{2}(9 - -9)(12) - \int_0^9 4x^{\frac{1}{2}}\,dx$ | M1 | Fully correct method for finding area of $R$: $\frac{1}{2}(\text{their } x_p - \text{``}-9\text{''})\!(\text{their } y_p) - \int_0^{\text{their } x_p} 4x^{\frac{1}{2}}\,dx$ |
| $\int 4x^{\frac{1}{2}}\,dx = \frac{4x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}(+c)$ or $\frac{8}{3}x^{\frac{3}{2}}(+c)$ | M1 | Integrates $\pm\lambda x^{\frac{1}{2}}$ to give $\pm\mu x^{\frac{3}{2}}$, where $\lambda, \mu \neq 0$ |
| $\frac{8}{3}x^{\frac{3}{2}}(+c)$ | A1 | Integrates $4x^{\frac{1}{2}}$ to give $\frac{8}{3}x^{\frac{3}{2}}$, simplified or unsimplified |
| $\text{Area}(R) = \frac{1}{2}(18)(12) - \frac{8}{3}\!\left(9^{\frac{3}{2}} - 0\right) = 108 - 72 = 36$ * | A1* | Fully correct proof leading to correct answer of 36 |
**(8 marks)**
---
## Part (c) — Alternative 1:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $B\!\left(-4,\frac{10}{3}\right)$ into $l \Rightarrow \frac{10p}{3} = -4 + 4p^2$ | M1 | Substitutes $x = -a$ and $y = \frac{10}{3}$ into $l$ |
| $6p^2 - 5p - 6 = 0 \Rightarrow (2p-3)(3p+2) = 0 \Rightarrow p = \ldots$ | M1 | Obtains 3-term quadratic and solves |
| $p = \frac{3}{2}$ into $l$ gives $\frac{3}{2}y = x + 4\!\left(\frac{3}{2}\right)^2 \Rightarrow x = \ldots$ | M1 | Substitutes their $p$ into $l$ and rearranges to give $x = \ldots$ |
| $x = \frac{3}{2}y - 9$ | A1 | |
| $p = \frac{3}{2} \Rightarrow P(9,12) \Rightarrow \text{Area}(R) = \int_0^{12}\!\left(\frac{1}{16}y^2 - \left(\frac{3}{2}y - 9\right)\right)dy$ | M1 | Fully correct method for area of $R$: $\int_0^{\text{their } y_p}\!\left(\frac{1}{16}y^2 - \text{their}\!\left(\frac{3}{2}y - 9\right)\right)dy$ |
| $\int\!\left(\frac{1}{16}y^2 - \frac{3}{2}y + 9\right)dy = \frac{1}{48}y^3 - \frac{3}{4}y^2 + 9y\,(+c)$ | M1, A1 | Integrates $\pm\lambda y^2 \pm \mu y \pm \nu$ to give $\pm\alpha y^3 \pm \beta y^2 \pm \nu y$; correct integration |
| $\text{Area}(R) = \left(\frac{1}{48}(12)^3 - \frac{3}{4}(12)^2 + 9(12)\right) - (0) = 36 - 108 + 108 = 36$ * | A1* | Fully correct proof leading to 36 |
**(8 marks)**
---
## Part (c) — Alternative 2:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $B\!\left(-4,\frac{10}{3}\right)$ into $l \Rightarrow \frac{10p}{3} = -4 + 4p^2$ | M1 | Substitutes $x = -a$ and $y = \frac{10}{3}$ into $l$ |
| $6p^2 - 5p - 6 = 0 \Rightarrow (2p-3)(3p+2) = 0 \Rightarrow p = \ldots$ | M1 | Obtains 3-term quadratic and solves |
| $p = \frac{3}{2}$ and $l$ cuts $x$-axis when $\frac{3}{2}(0) = x + 4\!\left(\frac{3}{2}\right)^2 \Rightarrow x = \ldots$ | M1 | Substitutes their $p$ (must be positive) and $y=0$ into $l$ |
| $x = -9$ | A1 | |
| $p = \frac{3}{2} \Rightarrow P(9,12)$ and $x=0$ in $l$: $y = \frac{2}{3}x + 6$ gives $y = 6$; $\Rightarrow \text{Area}(R) = \frac{1}{2}(9)(6) + \int_0^9\!\left(\left(\frac{2}{3}x + 6\right) - \left(4x^{\frac{1}{2}}\right)\right)dx$ | M1 | Fully correct method for area of $R$ |
| $\int\!\left(\frac{2}{3}x + 6 - 4x^{\frac{1}{2}}\right)dx = \frac{1}{3}x^2 + 6x - \frac{8}{3}x^{\frac{3}{2}}\,(+c)$ | M1, A1 | Integrates $\pm\lambda x \pm \mu \pm \nu x^{\frac{1}{2}}$ correctly |
| $\text{Area}(R) = 27 + \left(\left(\frac{1}{3}(9)^2 + 6(9) - \frac{8}{3}(9^{\frac{3}{2}})\right) - (0)\right) = 27 + (27 + 54 - 72) = 27 + 9 = 36$ * | A1* | Fully correct proof leading to 36 |
**(8 marks total for part c; 12 marks total for Question 5)**5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ff1fc9b0-6514-44e0-a2a3-46aa6411ce10-10_965_853_212_621}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Diagram not drawn to scale
$$\left[ Y \text { ou may quote without proof that for the general parabola } y ^ { 2 } = 4 a x , \frac { d y } { d x } = \frac { 2 a } { y } \right]$$
The parabola C has equation $\mathrm { y } ^ { 2 } = 16 \mathrm { x }$
\begin{enumerate}[label=(\alph*)]
\item Deduce that the point $\mathrm { P } \left( 4 \mathrm { p } ^ { 2 } , 8 \mathrm { p } \right)$ is a general point on C .
The line I is the tangent to C at the point P .
\item Show that an equation for I is
$$p y = x + 4 p ^ { 2 }$$
The finite region R , shown shaded in Figure 2, is bounded by the line I , the x -axis and the parabola C.\\
The line $I$ intersects the directrix of $C$ at the point $B$, where the $y$ coordinate of $B$ is $\frac { 10 } { 3 }$ Given that $\mathrm { p } > 0$
\item show that the area of R is 36
\section*{Q uestion 5 continued}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 AS Q5 [12]}}