AQA FP1 2007 January — Question 8 12 marks

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2007
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConic sections
TypeConic tangent through external point
DifficultyStandard +0.3 This is a straightforward hyperbola question testing standard techniques: substitution to find points, sketching, finding tangent equations, and determining tangency conditions. All parts follow routine procedures with no novel insight required. Part (d) involves algebraic manipulation and recognizing a repeated root indicates tangency, which is a standard A-level concept. Slightly easier than average due to the step-by-step guidance and familiar methods.
Spec1.02d Quadratic functions: graphs and discriminant conditions1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations1.03d Circles: equation (x-a)^2+(y-b)^2=r^2

8 A curve \(C\) has equation $$\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 9 } = 1$$
  1. Find the \(y\)-coordinates of the points on \(C\) for which \(x = 10\), giving each answer in the form \(k \sqrt { 3 }\), where \(k\) is an integer.
  2. Sketch the curve \(C\), indicating the coordinates of any points where the curve intersects the coordinate axes.
  3. Write down the equation of the tangent to \(C\) at the point where \(C\) intersects the positive \(x\)-axis.
    1. Show that, if the line \(y = x - 4\) intersects \(C\), the \(x\)-coordinates of the points of intersection must satisfy the equation $$16 x ^ { 2 } - 200 x + 625 = 0$$
    2. Solve this equation and hence state the relationship between the line \(y = x - 4\) and the curve \(C\).

Part (a)
AnswerMarks Guidance
\(x = 10 \Rightarrow 4 - \frac{y^2}{9} = 1\)M1
\(\Rightarrow y^2 = 27\)A1
\(\Rightarrow y = \pm 3\sqrt{3}\)A1 3 marks
Part (b)
AnswerMarks Guidance
One branch generally correctB1
Both branches correctB1
Intersections at \((\pm 5, 0)\)B1 3 marks
Part (c)
AnswerMarks Guidance
Required tangent is \(x = 5\)B1F 1 mark
Part (d)(i)
AnswerMarks Guidance
\(y\) correctly eliminatedM1
Fractions correctly clearedm1
\(16x^2 - 200x + 625 = 0\)A1 3 marks
Part (d)(ii)
AnswerMarks Guidance
\(x = \frac{25}{4}\)B1
Equal roots \(\Rightarrow\) tangencyE1 2 marks
Total for Question 8: 12 marks
GRAND TOTAL: 75 marks
### Part (a)
$x = 10 \Rightarrow 4 - \frac{y^2}{9} = 1$ | M1 | |
$\Rightarrow y^2 = 27$ | A1 | |
$\Rightarrow y = \pm 3\sqrt{3}$ | A1 | 3 marks | PI

### Part (b)
One branch generally correct | B1 | | Asymptotes not needed
Both branches correct | B1 | | With implied asymptotes
Intersections at $(\pm 5, 0)$ | B1 | 3 marks |

### Part (c)
Required tangent is $x = 5$ | B1F | 1 mark | ft wrong value in (b)

### Part (d)(i)
$y$ correctly eliminated | M1 | |
Fractions correctly cleared | m1 | |
$16x^2 - 200x + 625 = 0$ | A1 | 3 marks | convincingly shown (AG)

### Part (d)(ii)
$x = \frac{25}{4}$ | B1 | | No need to mention repeated root, but B0 if other values given as well
Equal roots $\Rightarrow$ tangency | E1 | 2 marks | Accept 'it's a tangent'

### **Total for Question 8: 12 marks**

---

## **GRAND TOTAL: 75 marks**
8 A curve $C$ has equation

$$\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 9 } = 1$$
\begin{enumerate}[label=(\alph*)]
\item Find the $y$-coordinates of the points on $C$ for which $x = 10$, giving each answer in the form $k \sqrt { 3 }$, where $k$ is an integer.
\item Sketch the curve $C$, indicating the coordinates of any points where the curve intersects the coordinate axes.
\item Write down the equation of the tangent to $C$ at the point where $C$ intersects the positive $x$-axis.
\item \begin{enumerate}[label=(\roman*)]
\item Show that, if the line $y = x - 4$ intersects $C$, the $x$-coordinates of the points of intersection must satisfy the equation

$$16 x ^ { 2 } - 200 x + 625 = 0$$
\item Solve this equation and hence state the relationship between the line $y = x - 4$ and the curve $C$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA FP1 2007 Q8 [12]}}