| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse locus problems |
| Difficulty | Challenging +1.8 This is a Further Maths FP1 question requiring knowledge of ellipse tangent equations, the chord of contact theorem, and algebraic manipulation. While the result is a standard theorem (chord of contact), students must recognize the relationship between the tangent intersection point and the chord equation, which requires conceptual understanding beyond routine calculation. The proof involves non-trivial coordinate geometry reasoning typical of harder Further Maths questions. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2 + 4y^2 = 4 \Rightarrow 2x + 8y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \ldots\) | M1 | 3.1a — Differentiates to obtain expression for \(\frac{dy}{dx}\) |
| Equation of tangent at \(P(x_1, y_1)\) is \((y - y_1) = -\frac{x_1}{4y_1}(x - x_1)\) | M1 | 3.1a — Forms general tangent equation at \((x_1, y_1)\) |
| \(xx_1 + 4yy_1 = x_1^2 + 4y_1^2 = 4\) and at \(Q(x_2, y_2)\): \(xx_2 + 4yy_2 = 4\) | A1 | 2.2a — Deduces correct equation and second equivalent |
| Intersect at \((r,s)\) gives \(rx_1 + 4sy_1 = 4\) and \(rx_2 + 4sy_2 = 4\) | B1 | 2.1 |
| Uses previous results to find gradient of line \(l\) | M1 | 3.1a |
| \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{-r}{4s}\) | A1 | 1.1b |
| Equation of \(l\) is \(y - y_1 = \frac{-r}{4s}(x - x_1)\) | M1 | 2.1 — Uses \(y - y_1 = m(x-x_1)\) or \(y = mx + c\) with gradient and attempts to find \(c\) |
| \(4sy + rx = 4sy_1 + rx_1 = 4\) | A1* | 2.2a — Correct solution; deduction that \(rx_1 + 4sy_1 = 4\) as \((x_1,y_1)\) on ellipse, no errors seen |
| Total: (8) | (8 marks) |
# Question 7:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 + 4y^2 = 4 \Rightarrow 2x + 8y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \ldots$ | M1 | 3.1a — Differentiates to obtain expression for $\frac{dy}{dx}$ |
| Equation of tangent at $P(x_1, y_1)$ is $(y - y_1) = -\frac{x_1}{4y_1}(x - x_1)$ | M1 | 3.1a — Forms general tangent equation at $(x_1, y_1)$ |
| $xx_1 + 4yy_1 = x_1^2 + 4y_1^2 = 4$ and at $Q(x_2, y_2)$: $xx_2 + 4yy_2 = 4$ | A1 | 2.2a — Deduces correct equation and second equivalent |
| Intersect at $(r,s)$ gives $rx_1 + 4sy_1 = 4$ and $rx_2 + 4sy_2 = 4$ | B1 | 2.1 |
| Uses previous results to find gradient of line $l$ | M1 | 3.1a |
| $\frac{y_2 - y_1}{x_2 - x_1} = \frac{-r}{4s}$ | A1 | 1.1b |
| Equation of $l$ is $y - y_1 = \frac{-r}{4s}(x - x_1)$ | M1 | 2.1 — Uses $y - y_1 = m(x-x_1)$ or $y = mx + c$ with gradient and attempts to find $c$ |
| $4sy + rx = 4sy_1 + rx_1 = 4$ | A1* | 2.2a — Correct solution; deduction that $rx_1 + 4sy_1 = 4$ as $(x_1,y_1)$ on ellipse, no errors seen |
| **Total: (8)** | | **(8 marks)** |\begin{enumerate}
\item $P$ and $Q$ are two distinct points on the ellipse described by the equation $x ^ { 2 } + 4 y ^ { 2 } = 4$
\end{enumerate}
The line $l$ passes through the point $P$ and the point $Q$.\\
The tangent to the ellipse at $P$ and the tangent to the ellipse at $Q$ intersect at the point $( r , s )$.\\
Show that an equation of the line $l$ is
$$4 s y + r x = 4$$
\hfill \mbox{\textit{Edexcel FP1 Q7 [8]}}