Edexcel IAL January 2024 P4 (WMA14/01) Q9 Parametric Equation, Differentiation, and Tangents
Figure 4 shows a sketch of the curve \( C \) with parametric equations
\(x = \sec t \quad y = \sqrt{3} \tan \left( t + \frac{\pi}{3} \right) \quad \frac{\pi}{6} < t < \frac{\pi}{2}\)
(a) Find \(\frac{dy}{dx}\) in terms of \( t \)(4)
(b) Find an equation for the tangent to \( C \) at the point where \( t = \frac{\pi}{3} \)
Give your answer in the form \( y = mx + c \), where \( m \) and \( c \) are constants.(4)
Give your answer in the form \( y = mx + c \), where \( m \) and \( c \) are constants.(4)
(c) Show that all point on \( C \) satisfy the equation
\(y = \frac{Ax^2 + B\sqrt{3}x^2 – 3}{4 – 3x^2}\)
Where \( A \) and \( B \) are constant to be found.(5)
Solution to Part a: Find \(\frac{dy}{dx}\) in terms of \( t \)
Key Concepts Used:
- Parametric differentiation: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)
- Trigonometric derivatives: \(\frac{d}{dt}(\sec t) = \sec t \tan t\), \(\frac{d}{dt}(\tan u) = \sec^2 u \cdot \frac{du}{dt}\)
- Simplification using trigonometric identities
Step-by-Step Working:
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Differentiate \(x\) and \(y\) with respect to \(t\)
\(x = \sec t \Rightarrow \frac{dx}{dt} = \sec t \tan t\)\(y = \sqrt{3}\tan\left(t + \frac{\pi}{3}\right) \Rightarrow \frac{dy}{dt} = \sqrt{3}\sec^2 \left(t + \frac{\pi}{3}\right)\)
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Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)
\(\frac{dy}{dx} = \frac{\sqrt{3}\sec^2 \left(t + \frac{\pi}{3}\right)}{\sec t \tan t}\)\(\frac{dy}{dx} = \sqrt{3} \cdot \frac{\sec^2 \left(t + \frac{\pi}{3}\right)}{\sec t \tan t}\)\(\frac{dy}{dx} = \sqrt{3} \cdot \frac{\sec \left(t + \frac{\pi}{3}\right)}{\tan t} \cdot \sec \left(t + \frac{\pi}{3}\right)\)
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Simplify using \(\sec\theta = \frac{1}{\cos\theta}, \tan\theta = \frac{\sin\theta}{\cos\theta}\)
\(\frac{dy}{dx} = \sqrt{3} \cdot \frac{1}{\cos \left(t + \frac{\pi}{3}\right)} \cdot \frac{\cos t}{\sin t} \cdot \frac{1}{\cos \left(t + \frac{\pi}{3}\right)}\)\(= \frac{\sqrt{3}\cos t}{\sin t \cos^2 \left(t + \frac{\pi}{3}\right)}\)\(= \sqrt{3}\cot t \sec^2 \left(t + \frac{\pi}{3}\right)\)
Final Answer
\(\frac{dy}{dx} = \sqrt{3}\cot t \sec^2 \left(t + \frac{\pi}{3}\right)\)
Solution to Part b: Tangent equation at \( t = \frac{\pi}{3} \)
Key Concepts Used:
- Parametric point evaluation: Substitute parameter value into \( x(t) \) and \( y(t) \)
- Tangent gradient: Evaluate \(\frac{dy}{dx}\) at given parameter value
- Straight line equation: \( y – y_1 = m(x – x_1) \)
Step-by-Step Working:
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Coordinates at \( t = \frac{\pi}{3} \)
\(x = \sec \left( \frac{\pi}{3} \right) = 2\)\(y = \sqrt{3} \tan \left( \frac{\pi}{3} + \frac{\pi}{3} \right) = \sqrt{3} \tan \left( \frac{2\pi}{3} \right)\)\(\tan \left( \frac{2\pi}{3} \right) = -\sqrt{3} \Rightarrow y = \sqrt{3} (-\sqrt{3}) = -3\)
So point is \((2, -3)\).
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Gradient at \( t = \frac{\pi}{3} \)
From part (a):\(\frac{dy}{dx} = \sqrt{3} \cot \left( \frac{\pi}{3} \right) \sec^2 \left( \frac{2\pi}{3} \right)\)\(\cot \left( \frac{\pi}{3} \right) = \frac{1}{\sqrt{3}}, \quad \sec \left( \frac{2\pi}{3} \right) = -2\)\(\Rightarrow \frac{dy}{dx} = \sqrt{3} \cdot \frac{1}{\sqrt{3}} \cdot 4 = 4\) -
Equation of tangent
\(y + 3 = 4(x – 2)\)\(y = 4x – 11\)
Final Answer
\(y = 4x – 11\)
Solution to Part c: Convert to Cartesian form \( y = \frac{Ax^2 + B\sqrt{3}x^2 – 3}{4 – 3x^2} \)
Key Concepts Used:
- Parametric elimination: Remove parameter t using \( \sec t = x \)
- Trigonometric identity: \( \sec^2t = 1 + \tan^2t \Rightarrow \tan t = \sqrt{x^2 – 1} \)
- Compound angle formula: \( \tan(A + B) = \frac{\tan A+\tan B}{1-\tan A\tan B} \)
- Algebraic manipulation: Rationalizing denominators and simplifying complex fractions
Step-by-Step Working:
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Use \( \sec t = x \Rightarrow \cos t = \frac{1}{x} \)
Also \( \tan\left(t + \frac{\pi}{3}\right) = \frac{\tan t+\sqrt{3}}{1-\sqrt{3}\tan t}\)
So:
\( y = \sqrt{3} \cdot \frac{\tan t + \sqrt{3}}{1 – \sqrt{3}\tan t}\) -
Express \( \tan t \) in terms of \( x \)
Since \( \sec^2t = 1 + \tan^2t \Rightarrow x^2 = 1 + \tan^2t \)\( \tan^2t = x^2 – 1 \Rightarrow \tan t = \sqrt{x^2 – 1} \quad (\text{since } t \in \left(\frac{\pi}{6}, \frac{\pi}{2}\right) \Rightarrow \tan t > 0)\) -
Substitute into \( y \)
\( y = \sqrt{3} \cdot \frac{\sqrt{x^2 – 1} + \sqrt{3}}{1 – \sqrt{3}\sqrt{x^2 – 1}}\)
Multiply numerator and denominator by \( 1 + \sqrt{3}\sqrt{x^2 – 1} \):
\( y = \sqrt{3} \cdot \frac{(\sqrt{x^2 – 1} + \sqrt{3})(1 + \sqrt{3}\sqrt{x^2 – 1})}{1 – 3(x^2 – 1)}\)Denominator: \( 1 – 3x^2 + 3 = 4 – 3x^2 \)
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Expand numerator
\( (\sqrt{x^2 – 1} + \sqrt{3})(1 + \sqrt{3}\sqrt{x^2 – 1}) \)
\( = \sqrt{x^2 – 1} + 3(x^2 – 1) + \sqrt{3} + 3\sqrt{x^2 – 1} \)
\( = 3x^2 – 3 + \sqrt{3} + 4\sqrt{x^2 – 1} \)
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Multiply by \(\sqrt{3}\):
\( \sqrt{3}(3x^2 – 3) + 3 + 4\sqrt{3}\sqrt{x^2 – 1} \)
\( = 3\sqrt{3}x^2 – 3\sqrt{3} + 3 + 4\sqrt{3}(x^2 – 1) \)
\( = 3\sqrt{3}x^2 + 3(1 – \sqrt{3}) + 4\sqrt{3}x^2 – 3 \)
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Identify A and B
Comparing with \( y = \frac{Ax^2 + B\sqrt{3}x^2 – 3}{4 – 3x^2} \):
\( A = 3\sqrt{3}, \quad B = 4 \)
Final Answer:
\( y = \frac{3\sqrt{3}x^2 + 4\sqrt{3}x^2 – 3}{4 – 3x^2} \)
So \( A = 3\sqrt{3}, \quad B = 4 \)
So \( A = 3\sqrt{3}, \quad B = 4 \)