| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2008 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Normal to curve at given point |
| Difficulty | Standard +0.3 This is a straightforward multi-part coordinate geometry question requiring standard techniques: finding a gradient by differentiation, using the perpendicular gradient property for the normal, then solving simultaneous equations. While it has multiple parts (7-8 marks total), each step follows routine procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.07m Tangents and normals: gradient and equations1.10c Magnitude and direction: of vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{8}{x^2}\) | B1 | Correct differentiation |
| \(m \text{ of tan} = 2\); \(m \text{ of normal} = -\frac{1}{2}\) | M1 | Use of \(m_1m_2 = -1\) |
| \(\text{Eqn of normal } y - 1 = -\frac{1}{2}(x - 2)\) | M1 | Correct method for line |
| \(\rightarrow 2y + x = 4\) | A1 | Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Sim eqns \(2y + x = 4, y = 5 - \frac{8}{x}\) | M1 | Complete elimination of \(x\) or \(y\) |
| \(\rightarrow x^2 + 6x - 16 = 0 \text{ or } y^2 - 7y + 6 = 0\) | DM1, A1 | Soln of quadratic. co |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Length} = \sqrt{10^2 + 5^2} = \sqrt{125}\) | M1 | Correct use of Pythagoras |
| \(\rightarrow 11.2 \text{ (accept } \sqrt{125} \text{ or } 5\sqrt{5} \text{ etc)}\) | A1 | For his points. |
$y = 5 - \frac{8}{x}, P(2, 1)$
**(i)**
$\frac{dy}{dx} = \frac{8}{x^2}$ | B1 | Correct differentiation
$m \text{ of tan} = 2$; $m \text{ of normal} = -\frac{1}{2}$ | M1 | Use of $m_1m_2 = -1$
$\text{Eqn of normal } y - 1 = -\frac{1}{2}(x - 2)$ | M1 | Correct method for line
$\rightarrow 2y + x = 4$ | A1 | Answer given
**(ii)**
Sim eqns $2y + x = 4, y = 5 - \frac{8}{x}$ | M1 | Complete elimination of $x$ or $y$
$\rightarrow x^2 + 6x - 16 = 0 \text{ or } y^2 - 7y + 6 = 0$ | DM1, A1 | Soln of quadratic. co
$\rightarrow (-8, 6)$
**(iii)**
$\text{Length} = \sqrt{10^2 + 5^2} = \sqrt{125}$ | M1 | Correct use of Pythagoras
$\rightarrow 11.2 \text{ (accept } \sqrt{125} \text{ or } 5\sqrt{5} \text{ etc)}$ | A1 | For his points.
---8 The equation of a curve is $y = 5 - \frac { 8 } { x }$.\\
(i) Show that the equation of the normal to the curve at the point $P ( 2,1 )$ is $2 y + x = 4$.
This normal meets the curve again at the point $Q$.\\
(ii) Find the coordinates of $Q$.\\
(iii) Find the length of $P Q$.
\hfill \mbox{\textit{CAIE P1 2008 Q8 [9]}}