| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Tangent to curve at given point |
| Difficulty | Standard +0.8 This is a structured multi-part question on parabola properties requiring coordinate geometry, calculus, and proof. Part (a) involves algebraic manipulation to find the chord equation (routine but multi-step). Parts (b) and (d) require showing specific geometric properties using the focus and perpendicularity, which demands understanding of parabola theory beyond rote application. While guided by parts, it requires sustained reasoning across multiple techniques, placing it moderately above average difficulty for Further Maths. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient \(= \frac{2ap - 2aq}{ap^2 - aq^2}\) seen | B1 | Correct statement for gradient |
| \(\frac{2ap-2aq}{ap^2-aq^2} = \frac{2a(p-q)}{a(p-q)(p+q)} = \frac{2}{p+q}\) or seen in equation | B1 | \(\frac{2}{p+q}\) can be seen later in solution |
| Uses \(y - y_1 = m(x-x_1)\) to give \((y-2aq) = \text{"m"}(x - aq^2)\) or equivalent with \(p\) | M1 | Use of correct formula through \(P\) or \(Q\); must be chord not tangent/normal |
| \((y-2aq) = \frac{2}{p+q}(x-aq^2)\) or \((y-2ap) = \frac{2}{p+q}(x-ap^2)\) or \(y = \frac{2}{p+q}x + \frac{2apq}{p+q}\) | A1 | Correct line equation with simplified gradient |
| See \(2aq^2\) or \(2ap^2\) term appear and disappear to give \(y(p+q) = 2x + 2apq\) | A1 cso | As given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \((a,0)\) into line equation to give \(0 = 2a + 2apq\) so \(pq = -1\) | B1 | For using \((a,0)\) to show \(pq=-1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 2a^{\frac{1}{2}}x^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = a^{\frac{1}{2}}x^{-\frac{1}{2}}\) or \(y^2=4ax \Rightarrow 2y\frac{dy}{dx}=4a\) or \(\frac{dy}{dx} = \frac{dy}{dp}\times\frac{dp}{dx} \Rightarrow \frac{dy}{dx} = 2a \times \frac{1}{2ap}\) | M1 | Use calculus to find \(dy/dx\); substitute coordinates of \(P\); may use chord gradient letting \(p \to q\) |
| So at \(P\) tangent gradient \(= \frac{1}{p}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(Q\) tangent gradient \(= \frac{1}{q}\) | B1 | \(1/q\) seen |
| \(\frac{1}{p} \times \frac{1}{q} = \frac{1}{pq} = \frac{1}{-1} = -1\), tangents are perpendicular or at right angles | B1 cso | \(\frac{1}{p}\times\frac{1}{q}=-1\) or \(\frac{1}{p}=-\frac{1}{1/q}\) or \(\frac{1}{q}=-\frac{1}{1/p}\); at least one intermediate step; words in bold with no errors |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient $= \frac{2ap - 2aq}{ap^2 - aq^2}$ seen | B1 | Correct statement for gradient |
| $\frac{2ap-2aq}{ap^2-aq^2} = \frac{2a(p-q)}{a(p-q)(p+q)} = \frac{2}{p+q}$ or seen in equation | B1 | $\frac{2}{p+q}$ can be seen later in solution |
| Uses $y - y_1 = m(x-x_1)$ to give $(y-2aq) = \text{"m"}(x - aq^2)$ or equivalent with $p$ | M1 | Use of correct formula through $P$ or $Q$; must be chord not tangent/normal |
| $(y-2aq) = \frac{2}{p+q}(x-aq^2)$ or $(y-2ap) = \frac{2}{p+q}(x-ap^2)$ or $y = \frac{2}{p+q}x + \frac{2apq}{p+q}$ | A1 | Correct line equation with simplified gradient |
| See $2aq^2$ or $2ap^2$ term appear and disappear to give $y(p+q) = 2x + 2apq$ | A1 cso | As given answer |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $(a,0)$ into line equation to give $0 = 2a + 2apq$ so $pq = -1$ | B1 | For using $(a,0)$ to show $pq=-1$ |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 2a^{\frac{1}{2}}x^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = a^{\frac{1}{2}}x^{-\frac{1}{2}}$ or $y^2=4ax \Rightarrow 2y\frac{dy}{dx}=4a$ or $\frac{dy}{dx} = \frac{dy}{dp}\times\frac{dp}{dx} \Rightarrow \frac{dy}{dx} = 2a \times \frac{1}{2ap}$ | M1 | Use calculus to find $dy/dx$; substitute coordinates of $P$; may use chord gradient letting $p \to q$ |
| So at $P$ tangent gradient $= \frac{1}{p}$ | A1 | |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $Q$ tangent gradient $= \frac{1}{q}$ | B1 | $1/q$ seen |
| $\frac{1}{p} \times \frac{1}{q} = \frac{1}{pq} = \frac{1}{-1} = -1$, tangents are **perpendicular** or at **right angles** | B1 cso | $\frac{1}{p}\times\frac{1}{q}=-1$ or $\frac{1}{p}=-\frac{1}{1/q}$ or $\frac{1}{q}=-\frac{1}{1/p}$; at least one intermediate step; words in bold with no errors |
---5. Points $P \left( a p ^ { 2 } , 2 a p \right)$ and $Q \left( a q ^ { 2 } , 2 a q \right)$, where $p ^ { 2 } \neq q ^ { 2 }$, lie on the parabola $y ^ { 2 } = 4 a x$.
\begin{enumerate}[label=(\alph*)]
\item Show that the chord $P Q$ has equation
$$y ( p + q ) = 2 x + 2 a p q$$
Given that this chord passes through the focus of the parabola,
\item show that $p q = - 1$
\item Using calculus find the gradient of the tangent to the parabola at $P$.
\item Show that the tangent to the parabola at $P$ and the tangent to the parabola at $Q$ are perpendicular.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2016 Q5 [10]}}