CAIE P1 2006 June — Question 5 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2006
SessionJune
Marks6
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Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypeLine and curve intersection
DifficultyStandard +0.3 This is a straightforward intersection problem requiring substitution of the linear equation into the parabola equation, solving the resulting quadratic, finding both points, and applying the distance formula. While it involves multiple steps, each is routine and the question follows a standard template with no conceptual challenges beyond basic algebraic manipulation.
Spec1.02c Simultaneous equations: two variables by elimination and substitution1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.10f Distance between points: using position vectors
5 The curve \(y ^ { 2 } = 12 x\) intersects the line \(3 y = 4 x + 6\) at two points. Find the distance between the two points.
AnswerMarks Guidance
\(y^2 = 12x\) and \(3y = 4x + 6\). Complete elimination of 1 variable. → \(y^2 - 9y + 18 = 0\) or \(4x^2 - 15x + 9 = 0\) solution → \((\frac{3}{4}, 3)\) and \((3, 6)\)M1 A1 DM1 A1 \(x\) or \(y\) must be removed completely. Must be a 3 term quad – not nec = 0. Correct method of solution. co.
Distance = \(\sqrt{(3^2 + 2.25^2)} = 3.75\)M1A1 Correct method including \(\sqrt{}\). co.
Total: [6]
$y^2 = 12x$ and $3y = 4x + 6$. Complete elimination of 1 variable. → $y^2 - 9y + 18 = 0$ or $4x^2 - 15x + 9 = 0$ solution → $(\frac{3}{4}, 3)$ and $(3, 6)$ | M1 A1 DM1 A1 | $x$ or $y$ must be removed completely. Must be a 3 term quad – not nec = 0. Correct method of solution. co.

Distance = $\sqrt{(3^2 + 2.25^2)} = 3.75$ | M1A1 | Correct method including $\sqrt{}$. co.

**Total: [6]**

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5 The curve $y ^ { 2 } = 12 x$ intersects the line $3 y = 4 x + 6$ at two points. Find the distance between the two points.

\hfill \mbox{\textit{CAIE P1 2006 Q5 [6]}}
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