| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola tangent/normal equation derivation |
| Difficulty | Challenging +1.2 This is a standard Further Pure 3 conic sections question requiring parametric differentiation to find the normal equation, then substitution at a specific angle. Part (a) involves routine implicit differentiation and algebraic manipulation. Part (b) requires finding the directrix equation (standard formula a²/c) and solving simultaneous equations. While it's Further Maths content making it inherently harder than Core, the techniques are well-practiced and methodical with no novel insights required. |
| Spec | 1.07s Parametric and implicit differentiation |
| VIIIV SIHI NI IIIUM IONOO | VI4V SIHI NI IM IMM ION OC | VI4V SIHI NI JIIYM IONOO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{3\sec^2\theta}{4\sec\theta\tan\theta}\left(=\frac{3}{4\sin\theta}\right)\) or implicit/explicit differentiation equivalent | M1 A1 | M1: Correct gradient method, finds \(\frac{dy}{d\theta}=p\sec^2\theta\) and \(\frac{dx}{d\theta}=q\sec\theta\tan\theta\). A1: Correct derivative in trig functions |
| Normal gradient \(= -\frac{4\sin\theta}{3}\) | M1 | Correct perpendicular gradient rule |
| \(y - 3\tan\theta = -\frac{4\sin\theta}{3}(x - 4\sec\theta)\) | M1 | Correct straight line method using gradient from calculus |
| \(3y + 4x\sin\theta = 25\tan\theta\) | A1* | Correct proof, no errors, one intermediate step required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b^2 = a^2(e^2-1) \Rightarrow 9=16(e^2-1) \Rightarrow e=\frac{5}{4}\) | M1 A1 | M1: Use correct eccentricity formula. A1: Correct value of \(e\), ignore \(\pm\) |
| \(x = \frac{a}{e} \Rightarrow x = \frac{16}{5}\) or \(\frac{4}{\frac{5}{4}}\) etc. | A1 | Correct value for \(\frac{a}{e}\), ignore \(\pm\) |
| \(\theta=\frac{\pi}{4}\), \(x=\frac{16}{5} \Rightarrow 3y + 2\sqrt{2}\times\frac{16}{5}=25\) | M1 | Substitutes \(\theta=\frac{\pi}{4}\) into normal equation using positive directrix |
| \(y = \frac{25}{3} - \frac{32}{15}\sqrt{2}\) | B1 B1 | B1: \(a=\frac{25}{3}\) oe or \(b=-\frac{32}{15}\) oe; B1: both correct |
# Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{3\sec^2\theta}{4\sec\theta\tan\theta}\left(=\frac{3}{4\sin\theta}\right)$ or implicit/explicit differentiation equivalent | M1 A1 | M1: Correct gradient method, finds $\frac{dy}{d\theta}=p\sec^2\theta$ and $\frac{dx}{d\theta}=q\sec\theta\tan\theta$. A1: Correct derivative in trig functions |
| Normal gradient $= -\frac{4\sin\theta}{3}$ | M1 | Correct perpendicular gradient rule |
| $y - 3\tan\theta = -\frac{4\sin\theta}{3}(x - 4\sec\theta)$ | M1 | Correct straight line method using gradient from calculus |
| $3y + 4x\sin\theta = 25\tan\theta$ | A1* | Correct proof, no errors, one intermediate step required |
**(5 marks)**
---
# Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b^2 = a^2(e^2-1) \Rightarrow 9=16(e^2-1) \Rightarrow e=\frac{5}{4}$ | M1 A1 | M1: Use correct eccentricity formula. A1: Correct value of $e$, ignore $\pm$ |
| $x = \frac{a}{e} \Rightarrow x = \frac{16}{5}$ or $\frac{4}{\frac{5}{4}}$ etc. | A1 | Correct value for $\frac{a}{e}$, ignore $\pm$ |
| $\theta=\frac{\pi}{4}$, $x=\frac{16}{5} \Rightarrow 3y + 2\sqrt{2}\times\frac{16}{5}=25$ | M1 | Substitutes $\theta=\frac{\pi}{4}$ into normal equation using positive directrix |
| $y = \frac{25}{3} - \frac{32}{15}\sqrt{2}$ | B1 B1 | B1: $a=\frac{25}{3}$ oe or $b=-\frac{32}{15}$ oe; B1: both correct |
Special case: If answer never seen in correct form but appears as $y=\frac{125-32\sqrt{2}}{15}$, allow B1B0.
**(6 marks)**
---5. The hyperbola $H$ has equation
$$\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 1$$
The point $P ( 4 \sec \theta , 3 \tan \theta ) , 0 < \theta < \frac { \pi } { 2 }$, lies on $H$.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation of the normal to $H$ at the point $P$ is
$$3 y + 4 x \sin \theta = 25 \tan \theta$$
The line $l$ is the directrix of $H$ for which $x > 0$\\
The normal to $H$ at $P$ crosses the line $l$ at the point $Q$. Given that $\theta = \frac { \pi } { 4 }$
\item find the $y$ coordinate of $Q$, giving your answer in the form $a + b \sqrt { 2 }$, where $a$ and $b$ are rational numbers to be found.\\
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VIIIV SIHI NI IIIUM IONOO & VI4V SIHI NI IM IMM ION OC & VI4V SIHI NI JIIYM IONOO \\
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\hfill \mbox{\textit{Edexcel FP3 2016 Q5 [11]}}