| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola tangent/normal equation derivation |
| Difficulty | Challenging +1.8 This question requires parametric differentiation on a hyperbola using the sec-tan parametrization, then algebraic manipulation to show the tangent passes through a focus with specific geometric properties. While the calculus is standard Further Maths fare, the connection between hyperbola tangent and ellipse focus, plus the geometric insight needed for part (b), elevates this beyond routine exercises. It's challenging but follows established Further Pure patterns. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{b^2}{a^2}\frac{a\sec\theta}{b\tan\theta}\left(=\frac{b}{a\sin\theta}\right)\) | M1A1 | M1: Attempts to differentiate implicitly or parametrically or directly. A1: Correct derivative as function of \(\theta\), any equivalent form |
| \(y - b\tan\theta = \frac{b}{a\sin\theta}(x-a\sec\theta)\) | dM1 | Correct straight line method, must be complete i.e. \(y=mx+c\) used, needs attempt at \(c\) |
| \(bx\sec\theta - ay\tan\theta = ab\) * | A1* cso | Reaches printed answer with at least one intermediate line of working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F\) is \((ae, 0)\) | B1 | Correct focus (if \(y\)-coordinate \(=0\) not used give B0) |
| \(abe\sec\theta = ab \Rightarrow e = \cos\theta\) | M1 | Substitute coordinates of focus into \(l\) |
| \(m = \frac{b}{a\sin\theta} = \frac{b}{a\sqrt{1-\cos^2\theta}} = \frac{b}{a\sqrt{1-e^2}}\) | M1 | Uses gradient of \(l\) to obtain expression for \(m\) in terms of \(a\), \(b\), \(e\) |
| For an ellipse \(b^2 = a^2(1-e^2)\) | M1 | Use of correct eccentricity formula for an ellipse in their expression for \(m\) |
| \(m = \frac{a\sqrt{1-e^2}}{a\sqrt{1-e^2}} = 1\) so \(l\) is parallel to \(y=x\) | A1 cso | Correct completion with no errors and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{b}{a\sin\theta}=1 \Rightarrow \sin\theta=\frac{b}{a}\) | B1 | \(\sin\theta=\frac{b}{a}\) |
| \(\sec\theta = \frac{1}{\cos\theta}=\frac{1}{\sqrt{1-\frac{b^2}{a^2}}}=\frac{a}{\sqrt{a^2-b^2}}\) | M1 | Attempt \(\sec\theta\) |
| \(bx\frac{a}{\sqrt{a^2-b^2}}-ay\cdot\frac{b}{...}=ab \Rightarrow x=\ldots \Rightarrow x=\sqrt{a^2-b^2}\) | M1 | Substitute for \(\sec\theta\), use \(y=0\), make \(x\) subject |
| For an ellipse \(b^2=a^2(1-e^2) \Rightarrow ae=\sqrt{a^2-b^2}\) | M1 | Use of correct eccentricity formula for an ellipse |
| So tangent passes through \((ae,0)\) which is \(F\) | A1 cso | Correct completion with no errors and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Focus is \((ae,0) = \left(\sqrt{a^2-b^2},0\right)\) | B1, M1 | B1: Focus correct. M1: Use of correct eccentricity formula |
| Line passes through the focus: \(b\sqrt{a^2-b^2}\sec\theta - 0 = ab\) | M1 | Line passes through the focus |
| \(\sec\theta=\frac{a}{\sqrt{a^2-b^2}} \Rightarrow \tan\theta=\frac{b}{\sqrt{a^2-b^2}}\) | M1 | Attempts \(\sec\theta\) and \(\tan\theta\) |
| \(x-y=\sqrt{a^2-b^2}\) OR sub \(\sec\theta\) and \(\tan\theta\) into gradient to get \(1\); \(\therefore\) parallel to \(y=x\) | A1 cso | Correct completion with no errors and conclusion |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{b^2}{a^2}\frac{a\sec\theta}{b\tan\theta}\left(=\frac{b}{a\sin\theta}\right)$ | M1A1 | M1: Attempts to differentiate implicitly or parametrically or directly. A1: Correct derivative as function of $\theta$, any equivalent form |
| $y - b\tan\theta = \frac{b}{a\sin\theta}(x-a\sec\theta)$ | dM1 | Correct straight line method, must be complete i.e. $y=mx+c$ used, needs attempt at $c$ |
| $bx\sec\theta - ay\tan\theta = ab$ * | A1* cso | Reaches printed answer with at least one intermediate line of working |
---
## Part (b) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F$ is $(ae, 0)$ | B1 | Correct focus (if $y$-coordinate $=0$ not used give B0) |
| $abe\sec\theta = ab \Rightarrow e = \cos\theta$ | M1 | Substitute coordinates of focus into $l$ |
| $m = \frac{b}{a\sin\theta} = \frac{b}{a\sqrt{1-\cos^2\theta}} = \frac{b}{a\sqrt{1-e^2}}$ | M1 | Uses gradient of $l$ to obtain expression for $m$ in terms of $a$, $b$, $e$ |
| For an ellipse $b^2 = a^2(1-e^2)$ | M1 | Use of correct eccentricity formula for an ellipse in their expression for $m$ |
| $m = \frac{a\sqrt{1-e^2}}{a\sqrt{1-e^2}} = 1$ so $l$ is parallel to $y=x$ | A1 cso | Correct completion with no errors and conclusion |
## Part (b) — Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{b}{a\sin\theta}=1 \Rightarrow \sin\theta=\frac{b}{a}$ | B1 | $\sin\theta=\frac{b}{a}$ |
| $\sec\theta = \frac{1}{\cos\theta}=\frac{1}{\sqrt{1-\frac{b^2}{a^2}}}=\frac{a}{\sqrt{a^2-b^2}}$ | M1 | Attempt $\sec\theta$ |
| $bx\frac{a}{\sqrt{a^2-b^2}}-ay\cdot\frac{b}{...}=ab \Rightarrow x=\ldots \Rightarrow x=\sqrt{a^2-b^2}$ | M1 | Substitute for $\sec\theta$, use $y=0$, make $x$ subject |
| For an ellipse $b^2=a^2(1-e^2) \Rightarrow ae=\sqrt{a^2-b^2}$ | M1 | Use of correct eccentricity formula for an ellipse |
| So tangent passes through $(ae,0)$ which is $F$ | A1 cso | Correct completion with no errors and conclusion |
## Part (b) — Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Focus is $(ae,0) = \left(\sqrt{a^2-b^2},0\right)$ | B1, M1 | B1: Focus correct. M1: Use of correct eccentricity formula |
| Line passes through the focus: $b\sqrt{a^2-b^2}\sec\theta - 0 = ab$ | M1 | Line passes through the focus |
| $\sec\theta=\frac{a}{\sqrt{a^2-b^2}} \Rightarrow \tan\theta=\frac{b}{\sqrt{a^2-b^2}}$ | M1 | Attempts $\sec\theta$ and $\tan\theta$ |
| $x-y=\sqrt{a^2-b^2}$ OR sub $\sec\theta$ and $\tan\theta$ into gradient to get $1$; $\therefore$ parallel to $y=x$ | A1 cso | Correct completion with no errors and conclusion |\begin{enumerate}
\item The hyperbola $H$ has equation $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$\\
and the ellipse $E$ has equation $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$\\
where $a > b > 0$\\
The line $l$ is a tangent to hyperbola $H$ at the point $P ( a \sec \theta , b \tan \theta )$, where $0 < \theta < \frac { \pi } { 2 }$\\
(a) Using calculus, show that an equation for $l$ is
\end{enumerate}
$$b x \sec \theta - a y \tan \theta = a b$$
Given that the point $F$ is the focus of ellipse $E$ for which $x > 0$ and that the line $l$ passes through $F$,\\
(b) show that $l$ is parallel to the line $y = x$
\hfill \mbox{\textit{Edexcel F3 2017 Q6 [9]}}