| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Rectangular hyperbola normal re-intersection |
| Difficulty | Standard +0.8 This is a Further Maths FP1 question requiring implicit differentiation to find the normal equation (part a), then solving a cubic equation when the normal re-intersects the hyperbola (part b). While the parametric form simplifies some work, finding the second intersection point requires algebraic manipulation and solving a cubic that factors using the known root, which is non-trivial for most students. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \tfrac{4}{x} = 4x^{-1} \Rightarrow \dfrac{dy}{dx} = -4x^{-2} = -\dfrac{4}{x^2}\) | M1 | Use of product rule: sum of two terms including \(dy/dx\), one of which is correct |
| \(xy = 4 \Rightarrow x\dfrac{dy}{dx} + y = 0\) | ||
| \(\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx} = -\dfrac{2}{t^2}\cdot\dfrac{1}{2}\) | their \(\dfrac{dy}{dt} \times \dfrac{1}{\text{their } \frac{dx}{dt}}\) | |
| \(\dfrac{dy}{dx} = -4x^{-2}\) or \(x\dfrac{dy}{dx}+y=0\) or \(\dfrac{dy}{dx}=-\dfrac{2}{t^2}\cdot\dfrac{1}{2}\) | A1 | Correct derivative \(-4x^{-2}\), \(-\dfrac{y}{x}\) or \(\dfrac{-1}{t^2}\) |
| \(m_N = t^2\) | M1 | Perpendicular gradient rule \(m_N m_T = -1\) |
| \(y - \dfrac{2}{t} = t^2(x - 2t)\) | M1 | \(y - \tfrac{2}{t} =\) their \(m_N(x-2t)\); gradient of normal must be different from tangent, from calculus, and a function of \(t\) |
| \(ty - t^3x = 2 - 2t^4\) * | A1* cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = -\tfrac{1}{2} \Rightarrow -\tfrac{1}{2}y - \left(-\tfrac{1}{2}\right)^3 x = 2 - 2\left(-\tfrac{1}{2}\right)^4\) | M1 | Substitutes the given value of \(t\) into the normal |
| \(4y - x + 15 = 0\) | ||
| \(y = \tfrac{4}{x} \Rightarrow x^2 - 15x - 16 = 0\) or \(\left(2t, \tfrac{2}{t}\right)\to\tfrac{8}{t}-2t+15=0\Rightarrow 2t^2-15t-8=0\) or \(x=\tfrac{4}{y}\Rightarrow 4y^2+15y-4=0\) | M1 | Substitutes to give a quadratic |
| \((x+1)(x-16)=0\) or \((2t+1)(t-8)=0\) or \((4y-1)(y+4)=0\) | M1 | Solves their 3TQ |
| \(P: x=-1,\ y=-4\) and \(Q: x=16,\ y=\tfrac{1}{4}\) | A1 | Correct values for \(x\) and \(y\) |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \tfrac{4}{x} = 4x^{-1} \Rightarrow \dfrac{dy}{dx} = -4x^{-2} = -\dfrac{4}{x^2}$ | M1 | Use of product rule: sum of two terms including $dy/dx$, one of which is correct |
| $xy = 4 \Rightarrow x\dfrac{dy}{dx} + y = 0$ | | |
| $\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx} = -\dfrac{2}{t^2}\cdot\dfrac{1}{2}$ | | their $\dfrac{dy}{dt} \times \dfrac{1}{\text{their } \frac{dx}{dt}}$ |
| $\dfrac{dy}{dx} = -4x^{-2}$ or $x\dfrac{dy}{dx}+y=0$ or $\dfrac{dy}{dx}=-\dfrac{2}{t^2}\cdot\dfrac{1}{2}$ | A1 | Correct derivative $-4x^{-2}$, $-\dfrac{y}{x}$ or $\dfrac{-1}{t^2}$ |
| $m_N = t^2$ | M1 | Perpendicular gradient rule $m_N m_T = -1$ |
| $y - \dfrac{2}{t} = t^2(x - 2t)$ | M1 | $y - \tfrac{2}{t} =$ their $m_N(x-2t)$; gradient of normal must be different from tangent, from calculus, and a function of $t$ |
| $ty - t^3x = 2 - 2t^4$ * | A1* cso | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = -\tfrac{1}{2} \Rightarrow -\tfrac{1}{2}y - \left(-\tfrac{1}{2}\right)^3 x = 2 - 2\left(-\tfrac{1}{2}\right)^4$ | M1 | Substitutes the given value of $t$ into the normal |
| $4y - x + 15 = 0$ | | |
| $y = \tfrac{4}{x} \Rightarrow x^2 - 15x - 16 = 0$ or $\left(2t, \tfrac{2}{t}\right)\to\tfrac{8}{t}-2t+15=0\Rightarrow 2t^2-15t-8=0$ or $x=\tfrac{4}{y}\Rightarrow 4y^2+15y-4=0$ | M1 | Substitutes to give a quadratic |
| $(x+1)(x-16)=0$ or $(2t+1)(t-8)=0$ or $(4y-1)(y+4)=0$ | M1 | Solves their 3TQ |
| $P: x=-1,\ y=-4$ and $Q: x=16,\ y=\tfrac{1}{4}$ | A1 | Correct values for $x$ and $y$ |
---4. The rectangular hyperbola $H$ has Cartesian equation $x y = 4$
The point $P \left( 2 t , \frac { 2 } { t } \right)$ lies on $H$, where $t \neq 0$
\begin{enumerate}[label=(\alph*)]
\item Show that an equation of the normal to $H$ at the point $P$ is
$$t y - t ^ { 3 } x = 2 - 2 t ^ { 4 }$$
The normal to $H$ at the point where $t = - \frac { 1 } { 2 }$ meets $H$ again at the point $Q$.
\item Find the coordinates of the point $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q4 [9]}}