| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola tangent intersection problems |
| Difficulty | Standard +0.8 This is a multi-part parabola question requiring implicit differentiation to find the tangent equation, then using geometric properties of parabolas (focus, directrix) to find intersection points and calculate an area. While the techniques are standard FP1 content, the question requires careful coordinate work across three connected parts and knowledge of parabola properties beyond just the equation, placing it moderately above average difficulty. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 2\sqrt{a}\,x^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \sqrt{a}\,x^{-\frac{1}{2}}\) or \(2y\frac{dy}{dx} = 4a\) | M1 | \(\frac{dy}{dx} = \pm k x^{-\frac{1}{2}}\) or \(ky\frac{dy}{dx} = c\) or their \(\frac{dy/dq}{dx/dq}\) |
| When \(x = aq^2\), \(m_T = \frac{\sqrt{a}}{\sqrt{a}q} = \frac{1}{q}\) | A1 | \(\frac{dy}{dx} = \frac{1}{q}\) |
| \(y - 2aq = \frac{1}{q}(x - aq^2)\); tangent: \(qy = x + aq^2\) | dM1; A1* | Applies \(y - 2aq = (m_T)(x - aq^2)\); cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(X\left(-\frac{1}{4}a, 0\right) \Rightarrow 0 = -\frac{1}{4}a + aq^2 \Rightarrow q = \frac{1}{2}\) (reject \(q = -\frac{1}{2}\)) | M1; A1 | Substitutes \(x = -\frac{1}{4}a\) and \(y=0\) into T |
| \(\frac{1}{2}y = -a + a\left(\frac{1}{2}\right)^2\); \(y = -\frac{3a}{2}\); \(D\left(-a, -\frac{3}{2}a\right)\) | M1; A1 | Substitutes \(q = \frac{1}{2}\) and \(x = -a\) in T or finds \(y_D = \frac{1}{q}(-a + aq^2)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Area}(FXD) = \frac{1}{2}\left(\frac{5a}{4}\right)\left(\frac{3a}{2}\right) = \frac{15a^2}{16}\) | M1; A1 cso | Applies \(\frac{1}{2}( |
| Answer | Marks | Guidance |
|---|---|---|
| Shoelace method with vertices \(\left(a, 0\right),\ \left(-\frac{1}{4}a, 0\right),\ \left(-a, -\frac{3}{2}a\right)\) | M1 | Correct attempt at shoelace |
| \(\frac{15a^2}{16}\) or \(0.9375a^2\) | A1 cao |
| Answer | Marks |
|---|---|
| Rectangle \(-\) triangle 1 \(-\) triangle 2 \(= 2a\cdot\frac{3a}{2} - \frac{1}{2}\cdot\frac{3a}{4}\cdot\frac{3a}{2} - \frac{1}{2}\cdot 2a\cdot\frac{3a}{2} = 3a^2 - \frac{9a^2}{16} - \frac{3a^2}{2}\) | M1 |
| \(\frac{15a^2}{16}\) or \(0.9375a^2\) | A1 cao |
| Answer | Marks |
|---|---|
| \(FX = \frac{5a}{4},\ FD = \frac{5a}{2},\ DX = \frac{3\sqrt{5}a}{4},\ \sin F = \frac{3}{5},\ \sin X = \frac{2}{\sqrt{5}}\); uses Area \(= \frac{1}{2}ab\sin C\) | M1 |
| \(\frac{15a^2}{16}\) or \(0.9375a^2\) | A1 cao |
# Question 7:
## Part (a):
| $y = 2\sqrt{a}\,x^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \sqrt{a}\,x^{-\frac{1}{2}}$ or $2y\frac{dy}{dx} = 4a$ | M1 | $\frac{dy}{dx} = \pm k x^{-\frac{1}{2}}$ or $ky\frac{dy}{dx} = c$ or their $\frac{dy/dq}{dx/dq}$ |
| When $x = aq^2$, $m_T = \frac{\sqrt{a}}{\sqrt{a}q} = \frac{1}{q}$ | A1 | $\frac{dy}{dx} = \frac{1}{q}$ |
| $y - 2aq = \frac{1}{q}(x - aq^2)$; tangent: $qy = x + aq^2$ | dM1; A1* | Applies $y - 2aq = (m_T)(x - aq^2)$; **cso** |
## Part (b):
| $X\left(-\frac{1}{4}a, 0\right) \Rightarrow 0 = -\frac{1}{4}a + aq^2 \Rightarrow q = \frac{1}{2}$ (reject $q = -\frac{1}{2}$) | M1; A1 | Substitutes $x = -\frac{1}{4}a$ and $y=0$ into **T** |
| $\frac{1}{2}y = -a + a\left(\frac{1}{2}\right)^2$; $y = -\frac{3a}{2}$; $D\left(-a, -\frac{3}{2}a\right)$ | M1; A1 | Substitutes $q = \frac{1}{2}$ and $x = -a$ in **T** or finds $y_D = \frac{1}{q}(-a + aq^2)$ |
## Part (c) - Way 1:
| $\text{Area}(FXD) = \frac{1}{2}\left(\frac{5a}{4}\right)\left(\frac{3a}{2}\right) = \frac{15a^2}{16}$ | M1; A1 cso | Applies $\frac{1}{2}(|FX|)(|y_D|)$; $\frac{15a^2}{16}$ or $0.9375a^2$ |
## Part (c) - Way 2:
| Shoelace method with vertices $\left(a, 0\right),\ \left(-\frac{1}{4}a, 0\right),\ \left(-a, -\frac{3}{2}a\right)$ | M1 | Correct attempt at shoelace |
| $\frac{15a^2}{16}$ or $0.9375a^2$ | A1 cao | |
## Part (c) - Way 3:
| Rectangle $-$ triangle 1 $-$ triangle 2 $= 2a\cdot\frac{3a}{2} - \frac{1}{2}\cdot\frac{3a}{4}\cdot\frac{3a}{2} - \frac{1}{2}\cdot 2a\cdot\frac{3a}{2} = 3a^2 - \frac{9a^2}{16} - \frac{3a^2}{2}$ | M1 | |
| $\frac{15a^2}{16}$ or $0.9375a^2$ | A1 cao | |
## Part (c) - Way 4:
| $FX = \frac{5a}{4},\ FD = \frac{5a}{2},\ DX = \frac{3\sqrt{5}a}{4},\ \sin F = \frac{3}{5},\ \sin X = \frac{2}{\sqrt{5}}$; uses Area $= \frac{1}{2}ab\sin C$ | M1 | |
| $\frac{15a^2}{16}$ or $0.9375a^2$ | A1 cao | |
---7. The parabola $C$ has equation $y ^ { 2 } = 4 a x$, where $a$ is a constant and $a > 0$ The point $Q \left( a q ^ { 2 } , 2 a q \right) , q > 0$, lies on the parabola $C$.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation of the tangent to $C$ at $Q$ is
$$q y = x + a q ^ { 2 }$$
The tangent to $C$ at the point $Q$ meets the $x$-axis at the point $X \left( - \frac { 1 } { 4 } a , 0 \right)$ and meets the directrix of $C$ at the point $D$.
\item Find, in terms of $a$, the coordinates of $D$.
Given that the point $F$ is the focus of the parabola $C$,
\item find the area, in terms of $a$, of the triangle $F X D$, giving your answer in its simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2017 Q7 [10]}}