Edexcel IAL October 2021 P4 (WMA14/01) Q1 Implicit Differentiation, Equation of Normal
The curve C has equation
\( 2x – 4y^2 + 3x^2y = 4x^2 + 8 \)
The point P (3, 2) lies on C.
Find the equation of the normal to C at the point P, writing your answer in the form ax + by + c = 0 where a, b and c are integers to be found.
(7)
Solution to Question: Normal to Curve at P(3, 2) for \( 2x – 4y^2 + 3x^2y = 4x^2 + 8 \)
Key Concepts Used:
- Implicit differentiation
- Normal line equation (negative reciprocal gradient)
Step-by-Step Working:
We need to find the equation of the normal line (a line perpendicular to the tangent at point P) to curve C at point P (3,2).
So, our approach would be first to find the derivative \( \frac{dy}{dx} \) (gradient of tangent) using implicit differentiation. Then, determine the gradient of the normal (negative reciprocal of tangent’s gradient). And at last, to use point-slope form to find the equation of the normal.
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Differentiate Each Term with Respect to x
We treat y as a function of x (y = y(x)) and apply the chain rule where necessary.First Term (2x):
\( \frac{d}{dx}(2x) = 2 \)Second Term (\( -4y^2 \)):
\( \frac{d}{dx}(-4y^2) = -8y\frac{dy}{dx} \)
(Chain rule: differentiate \( y^2 \) first to get 2y, then multiply by \( \frac{dy}{dx} \)) -
Third Term (\( 3x^2y \)):
\( \frac{d}{dx}(3x^2y) = 6xy + 3x^2\frac{dy}{dx} \)
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Right-Hand Side (4x² + 8):
\( \frac{d}{dx} (4x^2 + 8) = 8x \)
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Combined Differentiation:
\( 2 – 8y \frac{dy}{dx} + 6xy + 3x^2 \frac{dy}{dx} = 8x \)
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Solve for \( \frac{dy}{dx} \)
Group Terms Containing \( \frac{dy}{dx} \):
\( -8y \frac{dy}{dx} + 3x^2 \frac{dy}{dx} = 8x – 2 – 6xy \)Factor Out \( \frac{dy}{dx} \):
\( \frac{dy}{dx} (3x^2 – 8y) = 8x – 2 – 6xy \)Isolate \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{8x – 2 – 6xy}{3x^2 – 8y} \) -
Find Gradient of Tangent at P(3,2)
Substitute \( x = 3, y = 2 \):
\( \frac{dy}{dx} = \frac{8(3) – 2 – 6(3)(2)}{3(3)^2 – 8(2)} \)
\( \frac{dy}{dx} = \frac{24 – 2 – 36}{27 – 16} \)
\( \frac{dy}{dx} = \frac{-14}{11} \) -
Find Equation of the Normal Line, using Point-Slope Form
\( y-y_{1} = m(x-x_{1}) \)
Given \( P(3,2) \) and \( m = \frac{11}{14} \):
\( y-2 = \frac{11}{14}(x-3) \) -
Convert to Standard Form (\( ax+by+c=0 \))
Multiply both sides by 14 to eliminate fractions:
\( 14(y-2) = 11(x-3) \)
\( 14y-28 = 11x-33 \)
Rearrange terms:
\( 11x-14y-5 = 0 \)
(Product rule: \( \frac{d}{dx} (uv) = u’v + uv’ \), where \( u = 3x^2, v = y \))
The gradient of the tangent at \( P \) is \( -\frac{14}{11} \). The gradient of the normal is the negative reciprocal:
\( m_{normal} = \frac{11}{14} \)
Final Answer:
\( 11x-14y-5=0 \)
Verification: Mostly in these questions you may also crossly verify your answer, by substituting \( P(3,2) \) into the equation. If it comes out zero then it validates that line passes through it, or in other words, the coordinates satisfy the equation.
\( 11(3)-14(2)-5=33-28-5=0 \)
(Validates that the line passes through \( P \))
\( 11(3)-14(2)-5=33-28-5=0 \)
(Validates that the line passes through \( P \))