Edexcel C4 — Question 3 8 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks8
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TopicImplicit equations and differentiation
TypeFind normal equation at point
DifficultyStandard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate implicitly, substitute a point to find the gradient, then find the perpendicular gradient and write the normal equation. While it involves multiple steps (8 marks), each step follows a standard procedure taught in C4 with no novel problem-solving required, making it slightly easier than average.
Spec1.07s Parametric and implicit differentiation
3. A curve has the equation $$4 x ^ { 2 } - 2 x y - y ^ { 2 } + 11 = 0$$ Find an equation for the normal to the curve at the point with coordinates \(( - 1 , - 3 )\). (8)
3. continued
AnswerMarks Guidance
\(8x - 2y - 2x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0\)M1 A2
\((-1, -3) \Rightarrow -8 + 6 + 2 \frac{dy}{dx} + 6 \frac{dy}{dx} = 0, \quad \frac{dy}{dx} = \frac{1}{4}\)M1 A1
grad of normal \(= -4\)M1
\(\therefore y + 3 = -4(x + 1) \quad [y = -4x - 7]\)M1 A1 (8) 
$8x - 2y - 2x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0$ | M1 A2 |
$(-1, -3) \Rightarrow -8 + 6 + 2 \frac{dy}{dx} + 6 \frac{dy}{dx} = 0, \quad \frac{dy}{dx} = \frac{1}{4}$ | M1 A1 |
grad of normal $= -4$ | M1 |
$\therefore y + 3 = -4(x + 1) \quad [y = -4x - 7]$ | M1 A1 | **(8)**
3. A curve has the equation

$$4 x ^ { 2 } - 2 x y - y ^ { 2 } + 11 = 0$$

Find an equation for the normal to the curve at the point with coordinates $( - 1 , - 3 )$. (8)\\
3. continued\\

\hfill \mbox{\textit{Edexcel C4  Q3 [8]}}
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