| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Rectangular hyperbola tangent intersection |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question requiring implicit differentiation, parametric tangent equations, and locus finding. Part (a) is guided calculus verification, part (b) is routine application, but part (c) requires algebraic manipulation to eliminate the parameter and identify the locus form. While systematic, it demands more sophistication than typical A-level questions and involves FP1 content, placing it moderately above average difficulty. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations4.03d Linear transformations 2D: reflection, rotation, enlargement, shear |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y + x\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} = -\frac{\frac{-6}{t}}{\frac{1}{t^2}} = -\frac{1}{t^2} \cdot \frac{-6t^{-2}}{6} = -\frac{1}{t^2}\) | M1 | Differentiates implicitly, directly or parametrically to find gradient at \(P\) in terms of \(t\). Allow slips in coefficients |
| \(y - \frac{6}{t} = -\frac{1}{t^2}(x - 6t)\) | M1 | Finds equation of tangent at \(P\) using their gradient (not reciprocal). If using \(y=mx+c\) must proceed to find \(c\) |
| \(yt^2 + x = 12t\) | A1* | Correct equation for tangent at \(P\) from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = -\frac{y}{x} = \frac{-\frac{3}{t}}{12t} = -\frac{1}{4t^2}\) and \(y - \frac{3}{t} = -\frac{1}{4t^2}(x-12t)\) | M1 | Finds new gradient and proceeds to find equation of tangent at \(Q\). Alternatively replaces \(t\) by \(2t\) in answer to (a) |
| \(y - \frac{3}{t} = -\frac{1}{4t^2}(x-12t)\) o.e. such as \(4yt^2 + x = 24t\) | A1 | Correct equation, any form, need not be simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. \(\left.\begin{array}{l}4yt^2+x=24t\\yt^2+x=12t\end{array}\right\} 3yt^2=12t \Rightarrow y=\ldots\) and \(x=12t-yt^2=\ldots\) | M1 | Solves simultaneous equations to find both \(x\) and \(y\) coordinate for point \(R\) |
| \(x = 8t\) and \(y = \frac{4}{t}\) | A1 | Correct point of intersection, need not be simplified |
| \(xy = \ldots\) | dM1 | Dependent on first M1. Multiplies \(x\) by \(y\) to reach a constant |
| \(xy = 32\) hence rectangular hyperbola | A1 | Shows \(xy=32\) and hence rectangular hyperbola |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y + x\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} = -\frac{\frac{-6}{t}}{\frac{1}{t^2}} = -\frac{1}{t^2} \cdot \frac{-6t^{-2}}{6} = -\frac{1}{t^2}$ | M1 | Differentiates implicitly, directly or parametrically to find gradient at $P$ in terms of $t$. Allow slips in coefficients |
| $y - \frac{6}{t} = -\frac{1}{t^2}(x - 6t)$ | M1 | Finds equation of tangent at $P$ using their gradient (not reciprocal). If using $y=mx+c$ must proceed to find $c$ |
| $yt^2 + x = 12t$ | A1* | Correct equation for tangent at $P$ from correct working |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -\frac{y}{x} = \frac{-\frac{3}{t}}{12t} = -\frac{1}{4t^2}$ and $y - \frac{3}{t} = -\frac{1}{4t^2}(x-12t)$ | M1 | Finds new gradient and proceeds to find equation of tangent at $Q$. Alternatively replaces $t$ by $2t$ in answer to (a) |
| $y - \frac{3}{t} = -\frac{1}{4t^2}(x-12t)$ o.e. such as $4yt^2 + x = 24t$ | A1 | Correct equation, any form, need not be simplified |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. $\left.\begin{array}{l}4yt^2+x=24t\\yt^2+x=12t\end{array}\right\} 3yt^2=12t \Rightarrow y=\ldots$ and $x=12t-yt^2=\ldots$ | M1 | Solves simultaneous equations to find both $x$ and $y$ coordinate for point $R$ |
| $x = 8t$ and $y = \frac{4}{t}$ | A1 | Correct point of intersection, need not be simplified |
| $xy = \ldots$ | dM1 | Dependent on first M1. Multiplies $x$ by $y$ to reach a constant |
| $xy = 32$ hence **rectangular hyperbola** | A1 | Shows $xy=32$ and hence rectangular hyperbola |\begin{enumerate}
\item The rectangular hyperbola $H$ has equation $x y = 36$\\
(a) Use calculus to show that the equation of the tangent to $H$ at the point $P \left( 6 t , \frac { 6 } { t } \right)$ is
\end{enumerate}
$$y t ^ { 2 } + x = 12 t$$
The point $Q \left( 12 t , \frac { 3 } { t } \right)$ also lies on $H$.\\
(b) Find the equation of the tangent to $H$ at the point $Q$.
The tangent at $P$ and the tangent at $Q$ meet at the point $R$.\\
(c) Show that as $t$ varies the locus of $R$ is also a rectangular hyperbola.
\hfill \mbox{\textit{Edexcel FP1 2022 Q5 [9]}}