| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola locus problems |
| Difficulty | Challenging +1.2 This is a standard Further Maths hyperbola question requiring calculus to find the normal, coordinate geometry to find the midpoint, and elimination of the parameter to find a locus. While it involves multiple steps and Further Maths content (making it harder than typical A-level), the techniques are routine applications of standard methods with clear signposting through parts (a), (b), and (c). The parametric form is given, and each part builds systematically on the previous one without requiring novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\frac{dy}{dx}=\frac{b\sec^2\theta}{a\sec\theta\tan\theta}\) or \(\frac{b^2x}{a^2y}\) or \(\frac{bx}{a^2}\left(\frac{x^2}{a^2}-1\right)^{-\frac{1}{2}}\) | B1 | Correct tangent gradient in any form |
| \(m_N = -\frac{a\sec\theta\tan\theta}{b\sec^2\theta}\left(=-\frac{a}{b}\sin\theta\right)\) | M1 | Use parametric forms and correct perpendicular rule |
| \(y-b\tan\theta=-\frac{a}{b}\sin\theta(x-a\sec\theta)\) | M1A1 | M1: Correct straight line method using their \(m_N\); use of \(y=mx+c\) must include finding value for \(c\); A1: Correct equation any equivalent form |
| \(by-b^2\tan\theta=-ax\sin\theta+a^2\tan\theta\) | ||
| \(ax\sin\theta+by=(a^2+b^2)\tan\theta^*\) | A1* | Completes to printed answer with at least one intermediate step |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(y=0\Rightarrow x=\frac{(a^2+b^2)\tan\theta}{a\sin\theta}\left(=\frac{(a^2+b^2)}{a}\sec\theta\right)\) | B1 | Correct \(x\) coordinate |
| \(M\) is \(\left(\frac{1}{2}\left(\frac{a^2+b^2}{a}\sec\theta+a\sec\theta\right),\frac{b}{2}\tan\theta\right) = \left(\frac{2a^2+b^2}{2a}\sec\theta,\frac{b}{2}\tan\theta\right)\) oe | M1A1 | M1: Correct midpoint method for their \(x\) coordinate; A1: Correct coordinates for \(M\), any equivalent accepted; need not be in coordinate brackets |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\sec\theta=\frac{2ax}{2a^2+b^2},\ \tan\theta=\frac{2y}{b}\Rightarrow 1+\left(\frac{2y}{b}\right)^2=\left(\frac{2ax}{2a^2+b^2}\right)^2\) | M1A1 | M1: Correct attempt to eliminate \(\theta\) using coordinates of \(M\); A1: Correct equation |
| \(y^2=\frac{b^2}{4}\left(\frac{4a^2x^2}{(2a^2+b^2)^2}-1\right)\) oe | dM1A1 | dM1: Makes \(y^2\) the subject; A1: Correct equation in required form |
# Question 4:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\frac{dy}{dx}=\frac{b\sec^2\theta}{a\sec\theta\tan\theta}$ or $\frac{b^2x}{a^2y}$ or $\frac{bx}{a^2}\left(\frac{x^2}{a^2}-1\right)^{-\frac{1}{2}}$ | B1 | Correct tangent gradient in any form |
| $m_N = -\frac{a\sec\theta\tan\theta}{b\sec^2\theta}\left(=-\frac{a}{b}\sin\theta\right)$ | M1 | Use parametric forms and correct perpendicular rule |
| $y-b\tan\theta=-\frac{a}{b}\sin\theta(x-a\sec\theta)$ | M1A1 | M1: Correct straight line method using their $m_N$; use of $y=mx+c$ must include finding value for $c$; A1: Correct equation any equivalent form |
| $by-b^2\tan\theta=-ax\sin\theta+a^2\tan\theta$ | | |
| $ax\sin\theta+by=(a^2+b^2)\tan\theta^*$ | A1* | Completes to printed answer with at least one intermediate step |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $y=0\Rightarrow x=\frac{(a^2+b^2)\tan\theta}{a\sin\theta}\left(=\frac{(a^2+b^2)}{a}\sec\theta\right)$ | B1 | Correct $x$ coordinate |
| $M$ is $\left(\frac{1}{2}\left(\frac{a^2+b^2}{a}\sec\theta+a\sec\theta\right),\frac{b}{2}\tan\theta\right) = \left(\frac{2a^2+b^2}{2a}\sec\theta,\frac{b}{2}\tan\theta\right)$ oe | M1A1 | M1: Correct midpoint method for their $x$ coordinate; A1: Correct coordinates for $M$, any equivalent accepted; need not be in coordinate brackets |
## Part (c):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\sec\theta=\frac{2ax}{2a^2+b^2},\ \tan\theta=\frac{2y}{b}\Rightarrow 1+\left(\frac{2y}{b}\right)^2=\left(\frac{2ax}{2a^2+b^2}\right)^2$ | M1A1 | M1: Correct attempt to eliminate $\theta$ using coordinates of $M$; A1: Correct equation |
| $y^2=\frac{b^2}{4}\left(\frac{4a^2x^2}{(2a^2+b^2)^2}-1\right)$ oe | dM1A1 | dM1: Makes $y^2$ the subject; A1: Correct equation in required form |
**Total: 12 marks**
---4. The hyperbola $H$ has equation
$$\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$$
The line $l$ is a normal to $H$ at the point $P ( a \sec \theta , b \tan \theta ) , 0 < \theta < \frac { \pi } { 2 }$
\begin{enumerate}[label=(\alph*)]
\item Using calculus, show that an equation for $l$ is
$$a x \sin \theta + b y = \left( a ^ { 2 } + b ^ { 2 } \right) \tan \theta$$
The line $l$ meets the $x$-axis at the point $Q$, and the point $M$ is the midpoint of $P Q$.
\item Find the coordinates of $M$.
\item Hence find the cartesian equation of the locus of $M$ as $\theta$ varies, giving your answer in the form $y ^ { 2 } = \mathrm { f } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2018 Q4 [12]}}