Edexcel C4 — Question 2 8 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind normal equation at point
DifficultyStandard +0.8 This question requires implicit differentiation to find dy/dx, then calculating the gradient at a specific point, finding the perpendicular gradient for the normal, and converting to the required form. While the implicit differentiation itself is standard C4 content, the algebraic manipulation with multiple terms and the need to work with the normal (not tangent) adds moderate complexity beyond routine exercises.
Spec1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation
A curve has the equation $$3x^2 + xy - 2y^2 + 25 = 0.$$ Find an equation for the normal to the curve at the point with coordinates \((1, 4)\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. [8]
AnswerMarks Guidance
\(6x + y + x\frac{dy}{dx} - 4y\frac{dy}{dx} = 0\)M1 A2
\((1,4) \Rightarrow 6 + 4 + \frac{dy}{dx} - 16\frac{dy}{dx} = 0, \quad \frac{dy}{dx} = \frac{2}{3}\)M1 A1
grad of normal \(= -\frac{3}{2}\)M1
\(\therefore y - 4 = -\frac{3}{2}(x - 1)\)M1
\(2y - 8 = -3x + 3\)
\(3x + 2y - 11 = 0\)A1 (8) 
$6x + y + x\frac{dy}{dx} - 4y\frac{dy}{dx} = 0$ | M1 A2 |

$(1,4) \Rightarrow 6 + 4 + \frac{dy}{dx} - 16\frac{dy}{dx} = 0, \quad \frac{dy}{dx} = \frac{2}{3}$ | M1 A1 |

grad of normal $= -\frac{3}{2}$ | M1 |

$\therefore y - 4 = -\frac{3}{2}(x - 1)$ | M1 |

$2y - 8 = -3x + 3$ |
$3x + 2y - 11 = 0$ | A1 | (8) |

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A curve has the equation
$$3x^2 + xy - 2y^2 + 25 = 0.$$

Find an equation for the normal to the curve at the point with coordinates $(1, 4)$, giving your answer in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers. [8]

\hfill \mbox{\textit{Edexcel C4  Q2 [8]}}
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