| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola locus problems |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring parametric tangent equations to hyperbolas, coordinate geometry to find intersection points, locus derivation with parameter elimination, and inequality verification involving the focus. It demands multiple sophisticated techniques and careful algebraic manipulation across several steps, placing it well above average difficulty but not at the extreme end for FP1. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{x^2}{16} - \frac{y^2}{9} = 1 \Rightarrow \frac{x}{8} - \frac{2yy'}{9} = 0 \Rightarrow y' = \frac{9x}{16y} = \frac{36\cosh\theta}{48\sinh\theta}\) OR \(x = 4\cosh\theta, y = 3\sinh\theta \Rightarrow \frac{dy}{dx} = \frac{3\cosh\theta}{4\sinh\theta}\) | M1 | 3.1a |
| \(y - 3\sinh\theta = \frac{3\cosh\theta}{4\sinh\theta}(x - 4\cosh\theta)\) | M1 | 3.1a |
| \(y = 0 \Rightarrow x = \frac{4}{\cosh\theta}\) | A1 | 2.2a |
| Line \(l_2\) has equation \(x = 4\) | B1 | 2.2a |
| \(x = 4 \Rightarrow y - 3\sinh\theta = \frac{3\cosh\theta}{4\sinh\theta}(4 - 4\cosh\theta)\) | M1 | 2.1 |
| \(y = \frac{3\cosh\theta - 3}{\sinh\theta}\) | A1 | 2.2a |
| \(M\) is \(\left(\frac{1}{2}\left(4 + \frac{4}{\cosh\theta}\right), \frac{1}{2}\left(\frac{3\cosh\theta - 3}{\sinh\theta}\right)\right)\) | M1 | 1.1b |
| \(x = 2 + \frac{2}{\cosh\theta} \Rightarrow \cosh\theta = \frac{2}{x-2}\) then \(y^2 = \frac{9(\cosh\theta-1)^2}{4\sinh^2\theta} = \frac{9\left(\frac{2}{x-2}-1\right)^2}{4\left(\left(\frac{2}{x-2}\right)^2-1\right)}\) | M1 | 3.1a |
| \(= \frac{9\left(\frac{2}{x-2}-1\right)^2}{4\left(\frac{2}{x-2}-1\right)\left(\frac{2}{x-2}+1\right)} = \frac{9\left(\frac{2}{x-2}-1\right)}{4\left(\frac{2}{x-2}+1\right)} = \frac{9(4-x)}{4x}\) | A1* | 1.1b |
| \(p = 2\) or \(q = 4\) | M1 | 3.1a |
| \(p = 2\) and \(q = 4\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b^2 = a^2(e^2-1) \Rightarrow 9 = 16(e^2-1) \Rightarrow e = \frac{5}{4}\); Focus at \(x = ae = 4 \times \frac{5}{4} = 5\) | M1 | 1.1b |
| \(d >\) "5" \(- 4 = \ldots\) | M1 | 3.1a |
| \(d > 1\) | A1* | 1.1b |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{x^2}{16} - \frac{y^2}{9} = 1 \Rightarrow \frac{x}{8} - \frac{2yy'}{9} = 0 \Rightarrow y' = \frac{9x}{16y} = \frac{36\cosh\theta}{48\sinh\theta}$ OR $x = 4\cosh\theta, y = 3\sinh\theta \Rightarrow \frac{dy}{dx} = \frac{3\cosh\theta}{4\sinh\theta}$ | M1 | 3.1a | Attempts differentiation to obtain $\frac{dy}{dx}$ in terms of $\theta$. Allow for $\frac{x^2}{16} - \frac{y^2}{9} = 1 \Rightarrow \alpha x - \beta yy' = 0$ or parametric differentiation |
| $y - 3\sinh\theta = \frac{3\cosh\theta}{4\sinh\theta}(x - 4\cosh\theta)$ | M1 | 3.1a | Correct straight line method using coordinates of $P$ and gradient in terms of $\theta$ |
| $y = 0 \Rightarrow x = \frac{4}{\cosh\theta}$ | A1 | 2.2a | Uses $y=0$ to find $x$-coordinate of point $A$ |
| Line $l_2$ has equation $x = 4$ | B1 | 2.2a | Deduces equation of $l_2$ is $x=4$ |
| $x = 4 \Rightarrow y - 3\sinh\theta = \frac{3\cosh\theta}{4\sinh\theta}(4 - 4\cosh\theta)$ | M1 | 2.1 | Substitutes $x=4$ into tangent line equation to find point $B$ |
| $y = \frac{3\cosh\theta - 3}{\sinh\theta}$ | A1 | 2.2a | Correct $y$-coordinate of $B$ |
| $M$ is $\left(\frac{1}{2}\left(4 + \frac{4}{\cosh\theta}\right), \frac{1}{2}\left(\frac{3\cosh\theta - 3}{\sinh\theta}\right)\right)$ | M1 | 1.1b | Correct midpoint method for $AB$ |
| $x = 2 + \frac{2}{\cosh\theta} \Rightarrow \cosh\theta = \frac{2}{x-2}$ then $y^2 = \frac{9(\cosh\theta-1)^2}{4\sinh^2\theta} = \frac{9\left(\frac{2}{x-2}-1\right)^2}{4\left(\left(\frac{2}{x-2}\right)^2-1\right)}$ | M1 | 3.1a | Finds $\cosh\theta$ in terms of $x$ and substitutes to eliminate $\theta$ |
| $= \frac{9\left(\frac{2}{x-2}-1\right)^2}{4\left(\frac{2}{x-2}-1\right)\left(\frac{2}{x-2}+1\right)} = \frac{9\left(\frac{2}{x-2}-1\right)}{4\left(\frac{2}{x-2}+1\right)} = \frac{9(4-x)}{4x}$ | A1* | 1.1b | Obtains printed answer with no errors |
| $p = 2$ **or** $q = 4$ | M1 | 3.1a | |
| $p = 2$ **and** $q = 4$ | A1 | 1.1b | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b^2 = a^2(e^2-1) \Rightarrow 9 = 16(e^2-1) \Rightarrow e = \frac{5}{4}$; Focus at $x = ae = 4 \times \frac{5}{4} = 5$ | M1 | 1.1b | Complete method for finding $x$-coordinate of focus using correct eccentricity formula, finding $e$, then calculating $ae$ |
| $d >$ "5" $- 4 = \ldots$ | M1 | 3.1a | Subtracts 4 from $x$-coordinate of focus |
| $d > 1$ | A1* | 1.1b | Correct answer |\begin{enumerate}
\item The hyperbola $H$ has equation
\end{enumerate}
$$\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 1$$
The line $l _ { 1 }$ is the tangent to $H$ at the point $P ( 4 \cosh \theta , 3 \sinh \theta )$.\\
The line $l _ { 1 }$ meets the $x$-axis at the point $A$.\\
The line $l _ { 2 }$ is the tangent to $H$ at the point $( 4,0 )$.\\
The lines $l _ { 1 }$ and $l _ { 2 }$ meet at the point $B$ and the midpoint of $A B$ is the point $M$.\\
(a) Show that, as $\theta$ varies, a Cartesian equation for the locus of $M$ is
$$y ^ { 2 } = \frac { 9 ( 4 - x ) } { 4 x } \quad p < x < q$$
where $p$ and $q$ are values to be determined.
Let $S$ be the focus of $H$ that lies on the positive $x$-axis.\\
(b) Show that the distance from $M$ to $S$ is greater than 1
\hfill \mbox{\textit{Edexcel FP1 2019 Q8 [14]}}