| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic tangent through external point |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on hyperbola tangents with clear scaffolding. Part (a) is routine substitution and algebra, part (b) applies the discriminant condition for tangency (standard technique), and part (c) requires solving a quadratic for m then finding c. While it's Further Maths content (inherently harder), the question guides students through each step methodically with no novel geometric insight required, making it moderately above average difficulty. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{x^2}{a^2} - \frac{(mx+c)^2}{b^2} = 1\) and so \(b^2x^2 - a^2(mx+c)^2 = a^2b^2\) | M1 | |
| \(\therefore (b^2-a^2m^2)x^2 - 2a^2mcx - a^2(c^2+b^2) = 0\) or \((a^2m^2-b^2)x^2 + 2a^2mcx + a^2(c^2+b^2) = 0\) ✱ | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((2a^2mc)^2 = 4(a^2m^2-b^2)\times a^2(c^2+b^2)\) | M1 | |
| \(4a^4m^2c^2 = -4a^2(b^2c^2+b^4-a^2m^2c^2-a^2m^2b^2)\) | ||
| \(c^2 = a^2m^2 - b^2\) or \(a^2m^2 = b^2+c^2\) ✱ | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitute \((1,4)\) into \(y=mx+c\) to give \(4=m+c\) | B1 | |
| Substitute \(a=5\) and \(b=4\) into \(c^2=a^2m^2-b^2\) to give \(c^2=25m^2-16\) | M1 | |
| Solve simultaneous equations to eliminate \(m\) or \(c\): \((4-m)^2=25m^2-16\) | A1 | |
| To obtain \(24m^2+8m-32=0\) | M1 | |
| Solve to obtain \(8(3m+4)(m-1)=0\ldots m=\ldots\) or... \(m=1\) or \(-\frac{4}{3}\) | A1 | |
| Substitute to get \(c=3\) or \(\frac{16}{3}\) | M1 | |
| Lines are \(y=x+3\) and \(3y+4x=16\) | A1 | (7) [11] |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{x^2}{a^2} - \frac{(mx+c)^2}{b^2} = 1$ and so $b^2x^2 - a^2(mx+c)^2 = a^2b^2$ | M1 | |
| $\therefore (b^2-a^2m^2)x^2 - 2a^2mcx - a^2(c^2+b^2) = 0$ or $(a^2m^2-b^2)x^2 + 2a^2mcx + a^2(c^2+b^2) = 0$ ✱ | A1 | (2) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2a^2mc)^2 = 4(a^2m^2-b^2)\times a^2(c^2+b^2)$ | M1 | |
| $4a^4m^2c^2 = -4a^2(b^2c^2+b^4-a^2m^2c^2-a^2m^2b^2)$ | | |
| $c^2 = a^2m^2 - b^2$ or $a^2m^2 = b^2+c^2$ ✱ | A1 | (2) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $(1,4)$ into $y=mx+c$ to give $4=m+c$ | B1 | |
| Substitute $a=5$ and $b=4$ into $c^2=a^2m^2-b^2$ to give $c^2=25m^2-16$ | M1 | |
| Solve simultaneous equations to eliminate $m$ or $c$: $(4-m)^2=25m^2-16$ | A1 | |
| To obtain $24m^2+8m-32=0$ | M1 | |
| Solve to obtain $8(3m+4)(m-1)=0\ldots m=\ldots$ or... $m=1$ or $-\frac{4}{3}$ | A1 | |
| Substitute to get $c=3$ or $\frac{16}{3}$ | M1 | |
| Lines are $y=x+3$ and $3y+4x=16$ | A1 | (7) **[11]** |
---\begin{enumerate}
\item The hyperbola $H$ has equation $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$, where $a$ and $b$ are constants.
\end{enumerate}
The line $L$ has equation $y = m x + c$, where $m$ and $c$ are constants.\\
(a) Given that $L$ and $H$ meet, show that the $x$-coordinates of the points of intersection are the roots of the equation
$$\left( a ^ { 2 } m ^ { 2 } - b ^ { 2 } \right) x ^ { 2 } + 2 a ^ { 2 } m c x + a ^ { 2 } \left( c ^ { 2 } + b ^ { 2 } \right) = 0$$
Hence, given that $L$ is a tangent to $H$,\\
(b) show that $a ^ { 2 } m ^ { 2 } = b ^ { 2 } + c ^ { 2 }$.
The hyperbola $H ^ { \prime }$ has equation $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 16 } = 1$.\\
(c) Find the equations of the tangents to $H ^ { \prime }$ which pass through the point $( 1,4 )$.\\
\hfill \mbox{\textit{Edexcel FP3 2009 Q6 [11]}}