| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic tangent through external point |
| Difficulty | Challenging +1.2 This is a standard Further Maths conic sections question requiring substitution of a line into an ellipse equation and using the discriminant condition for tangency (part a), then solving a system to find specific tangents through a point (part b). While it involves multiple steps and algebraic manipulation, the technique is routine for F3 students and follows a well-established method with no novel insight required. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Tangent at \((5\cos\theta, 3\sin\theta)\) is \(y = -\frac{3\cos\theta}{5\sin\theta}x + \frac{3}{\sin\theta}\) | M1A1 | M1: Full attempt at a general tangent; A1: Correct tangent |
| \(c^2 - 25m^2 = \frac{9}{\sin^2\theta} - 25\cdot\frac{9\cos^2\theta}{25\sin^2\theta}\) | M1 | Substitutes their \(c\) and \(m\) into \(c^2-25m^2\) |
| \(c^2 - 25m^2 = 9\) * | A1* | Achieves printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{x^2}{25}+\frac{y^2}{9}=1 \Rightarrow \frac{2x}{25}+\frac{2y}{9}\frac{dy}{dx}=0\) | ||
| \(\frac{dy}{dx} = -\frac{9x}{25y} \Rightarrow m = -\frac{9x}{25y}\) | ||
| \(y = -\frac{9x}{25y} = mx+c \Rightarrow x = -\frac{25mc}{9+25m^2}\) and \(y = \frac{9c}{9+25m^2}\) | M1A1 | M1: Differentiates implicitly, uses \(\frac{dy}{dx}=m\) and \(y=mx+c\) to obtain \(x\) and \(y\) in terms of \(m\) and \(c\); A1: Correct \(x\) and \(y\) |
| \(\frac{25m^2c^2}{(9+25m^2)^2}+\frac{9c^2}{(9+25m^2)^2}=1\) | M1 | Substitutes their \(x\) and \(y\) in terms of \(m\) and \(c\) into \(E\) |
| \(25m^2c^2+9c^2=(9+25m^2)^2 \Rightarrow c^2-25m^2=9\) * | A1* | Achieves printed answer with no errors |
## Question 5(a) Alternatives:
**Way 2:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Tangent at $(5\cos\theta, 3\sin\theta)$ is $y = -\frac{3\cos\theta}{5\sin\theta}x + \frac{3}{\sin\theta}$ | M1A1 | M1: Full attempt at a general tangent; A1: Correct tangent |
| $c^2 - 25m^2 = \frac{9}{\sin^2\theta} - 25\cdot\frac{9\cos^2\theta}{25\sin^2\theta}$ | M1 | Substitutes their $c$ and $m$ into $c^2-25m^2$ |
| $c^2 - 25m^2 = 9$ * | A1* | Achieves printed answer with no errors |
**Way 3:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{x^2}{25}+\frac{y^2}{9}=1 \Rightarrow \frac{2x}{25}+\frac{2y}{9}\frac{dy}{dx}=0$ | | |
| $\frac{dy}{dx} = -\frac{9x}{25y} \Rightarrow m = -\frac{9x}{25y}$ | | |
| $y = -\frac{9x}{25y} = mx+c \Rightarrow x = -\frac{25mc}{9+25m^2}$ and $y = \frac{9c}{9+25m^2}$ | M1A1 | M1: Differentiates implicitly, uses $\frac{dy}{dx}=m$ and $y=mx+c$ to obtain $x$ and $y$ in terms of $m$ and $c$; A1: Correct $x$ and $y$ |
| $\frac{25m^2c^2}{(9+25m^2)^2}+\frac{9c^2}{(9+25m^2)^2}=1$ | M1 | Substitutes their $x$ and $y$ in terms of $m$ and $c$ into $E$ |
| $25m^2c^2+9c^2=(9+25m^2)^2 \Rightarrow c^2-25m^2=9$ * | A1* | Achieves printed answer with no errors |
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\item The ellipse $E$ has equation $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1$
\end{enumerate}
The line $L$ has equation $y = m x + c$, where $m$ and $c$ are constants.\\
Given that $L$ is a tangent to $E$,\\
(a) show that
$$c ^ { 2 } - 25 m ^ { 2 } = 9$$
(b) find the equations of the tangents to $E$ which pass through the point $( 3,4 )$.\\
\hfill \mbox{\textit{Edexcel F3 2015 Q5 [9]}}