Edexcel C4 2005 June — Question 2 7 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2005
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind stationary points
DifficultyStandard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate implicitly, set dy/dx = 0, and solve the resulting system. While it involves multiple steps (implicit differentiation, algebraic manipulation, substitution back into the original equation), each step uses standard C4 techniques with no novel insight required. It's slightly easier than average because the algebra is manageable and the method is routine for this topic.
Spec1.07s Parametric and implicit differentiation
2. A curve has equation $$x ^ { 2 } + 2 x y - 3 y ^ { 2 } + 16 = 0 .$$ Find the coordinates of the points on the curve where \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\).
AnswerMarks
\(2x + \left(2x\left(\frac{dy}{dx}+2y\right)\right) - 6y\frac{dy}{dx} = 0\)M1 (A1) A1
\(\frac{dy}{dx} = 0 \Rightarrow x + y = 0\) or equivalentM1
Eliminating either variable and solving for at least one value of \(x\) or \(y\): \(y^2 - 2y^2 - 3y^2 + 16 = 0\) or the same equation in \(x\); \(y = \pm 2\) or \(x = \pm 2\)M1
\((2, -2), (-2, 2)\)A1, A1
Note: \(\frac{dy}{dx} = \frac{x+y}{3y-x}\)
Alternative:
AnswerMarks
\(3y^2 - 2xy - \left(x^2 + 16\right) = 0\)
\(y = \frac{2x \pm \sqrt{16x^2 + 192}}{6}\)
\(\frac{dy}{dx} = \frac{1}{3} \pm \frac{1}{3}\cdot\frac{8x}{\sqrt{16x^2+192}}\)M1 A1 ± A1
\(\frac{dy}{dx} = 0 \Rightarrow \frac{8x}{\sqrt{16x^2+192}} = \pm 1\)M1
\(64x^2 = 16x^2 + 192\)M1 A1
\(x = \pm 2\)
\((2, -2), (-2, 2)\)A1
Total: [7]
$2x + \left(2x\left(\frac{dy}{dx}+2y\right)\right) - 6y\frac{dy}{dx} = 0$ | M1 (A1) A1 |

$\frac{dy}{dx} = 0 \Rightarrow x + y = 0$ or equivalent | M1 |

Eliminating either variable and solving for at least one value of $x$ or $y$: $y^2 - 2y^2 - 3y^2 + 16 = 0$ or the same equation in $x$; $y = \pm 2$ or $x = \pm 2$ | M1 |

$(2, -2), (-2, 2)$ | A1, A1 |

Note: $\frac{dy}{dx} = \frac{x+y}{3y-x}$ | |

**Alternative:**

$3y^2 - 2xy - \left(x^2 + 16\right) = 0$ | |

$y = \frac{2x \pm \sqrt{16x^2 + 192}}{6}$ | |

$\frac{dy}{dx} = \frac{1}{3} \pm \frac{1}{3}\cdot\frac{8x}{\sqrt{16x^2+192}}$ | M1 A1 ± A1 |

$\frac{dy}{dx} = 0 \Rightarrow \frac{8x}{\sqrt{16x^2+192}} = \pm 1$ | M1 |

$64x^2 = 16x^2 + 192$ | M1 A1 |

$x = \pm 2$ | |

$(2, -2), (-2, 2)$ | A1 |

**Total: [7]**

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2. A curve has equation

$$x ^ { 2 } + 2 x y - 3 y ^ { 2 } + 16 = 0 .$$

Find the coordinates of the points on the curve where $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$.\\

\hfill \mbox{\textit{Edexcel C4 2005 Q2 [7]}}
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