Edexcel FP1 2014 June — Question 8 5 marks

Exam BoardEdexcel
ModuleFP1 (Further Pure Mathematics 1)
Year2014
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConic sections
TypeRectangular hyperbola tangent intersection
DifficultyChallenging +1.2 This is a standard FP1 rectangular hyperbola problem requiring students to use the given tangent formula with two unknowns, set up simultaneous equations by substituting the intersection point, and solve for the parameters. While it involves multiple steps and algebraic manipulation, it follows a predictable pattern typical of FP1 conic sections questions with no novel insight required.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations

8. The rectangular hyperbola \(H\) has equation \(x y = c ^ { 2 }\), where \(c\) is a positive constant. The point \(P \left( c t , \frac { c } { t } \right) , t \neq 0\), is a general point on \(H\). An equation for the tangent to \(H\) at \(P\) is given by $$y = - \frac { 1 } { t ^ { 2 } } x + \frac { 2 c } { t }$$ The points \(A\) and \(B\) lie on \(H\).
The tangent to \(H\) at \(A\) and the tangent to \(H\) at \(B\) meet at the point \(\left( - \frac { 6 } { 7 } c , \frac { 12 } { 7 } c \right)\).
Find, in terms of \(c\), the coordinates of \(A\) and the coordinates of \(B\).

Question 8:
AnswerMarks Guidance
AnswerMark Guidance
\(\frac{12}{7}c = -\frac{1}{t^2} \times -\frac{6}{7}c + \frac{2c}{t}\)M1 Substitutes \(\left(-\frac{6}{7}c,\ \frac{12}{7}c\right)\) into the equation of the tangent
\(\frac{12}{7}c = -\frac{1}{t^2}\cdot\left(-\frac{6}{7}c\right) + \frac{2c}{t} \Rightarrow 6t^2 - 7t - 3 = 0\)A1 Correct 3TQ in terms of \(t\)
\(6t^2 - 7t - 3 = 0 \Rightarrow (3t+1)(2t-3) = 0 \Rightarrow t =\)M1 Attempt to solve their 3TQ for \(t\)
\(t = -\frac{1}{3},\ t = \frac{3}{2} \Rightarrow \left(-\frac{1}{3}c,\ -3c\right),\ \left(\frac{3}{2}c,\ \frac{2}{3}c\right)\)M1A1 M1: Uses at least one of their values of \(t\) to find \(A\) or \(B\); A1: Correct coordinates
(Total: 5)
## Question 8:

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{12}{7}c = -\frac{1}{t^2} \times -\frac{6}{7}c + \frac{2c}{t}$ | M1 | Substitutes $\left(-\frac{6}{7}c,\ \frac{12}{7}c\right)$ into the equation of the tangent |
| $\frac{12}{7}c = -\frac{1}{t^2}\cdot\left(-\frac{6}{7}c\right) + \frac{2c}{t} \Rightarrow 6t^2 - 7t - 3 = 0$ | A1 | Correct 3TQ in terms of $t$ |
| $6t^2 - 7t - 3 = 0 \Rightarrow (3t+1)(2t-3) = 0 \Rightarrow t =$ | M1 | Attempt to solve their 3TQ for $t$ |
| $t = -\frac{1}{3},\ t = \frac{3}{2} \Rightarrow \left(-\frac{1}{3}c,\ -3c\right),\ \left(\frac{3}{2}c,\ \frac{2}{3}c\right)$ | M1A1 | M1: Uses at least one of their values of $t$ to find $A$ or $B$; A1: Correct coordinates |

**(Total: 5)**

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8. The rectangular hyperbola $H$ has equation $x y = c ^ { 2 }$, where $c$ is a positive constant. The point $P \left( c t , \frac { c } { t } \right) , t \neq 0$, is a general point on $H$.

An equation for the tangent to $H$ at $P$ is given by

$$y = - \frac { 1 } { t ^ { 2 } } x + \frac { 2 c } { t }$$

The points $A$ and $B$ lie on $H$.\\
The tangent to $H$ at $A$ and the tangent to $H$ at $B$ meet at the point $\left( - \frac { 6 } { 7 } c , \frac { 12 } { 7 } c \right)$.\\
Find, in terms of $c$, the coordinates of $A$ and the coordinates of $B$.\\

\hfill \mbox{\textit{Edexcel FP1 2014 Q8 [5]}}