Edexcel IAL June 2022 P4 (WMA14/01) Q7 Parametric Equations, Equations of Tangents
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
The curve \( C \) has parametric equations
\( x = \sin t – 3 \cos^2 t \)
\( y = 3 \sin t + 2 \cos t \)
\( 0 \leq t \leq 5 \)
\( y = 3 \sin t + 2 \cos t \)
\( 0 \leq t \leq 5 \)
(a) Show that \(\frac{dy}{dx} = 3\) where \(t = \pi\)(4)
The point P lies on \(C\) where \(t = \pi\)
(b) Find the equation of the tangent to the curve at \(P\) in the form \(y = mx + c\) where \(m\) and \(c\) are constants to be found.(2)
Given that the tangent to the curve at \(P\) cuts \(C\) at the point \(Q\)
(c) show that the value of \(t\) at point \(Q\) satisfies the equation
\( 9 \cos^2 t + 2 \cos t – 7 = 0 \)
(3)
(d) Hence find the exact value of the \(y\) coordinate of \(Q\)(3)
Solution to Part a: Showing \(\frac{dy}{dx} = 3\) at \(t = \pi\)
Key Concepts Used:
- Parametric Differentiation
Step-by-Step Working:
Verify the derivative condition using parametric differentiation.
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Compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
For \(x = \sin t – 3 \cos^2 t\):
\( \frac{dx}{dt} = \cos t + 6 \cos t \sin t \)
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Find \(\frac{dy}{dx}\)
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3\cos t – 2\sin t}{\cos t + 6\cos t\sin t} \)
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Evaluate at \( t = \pi \)
\( \frac{dy}{dx} = \frac{3\cos\pi – 2\sin\pi}{\cos\pi + 6\cos\pi\sin\pi} = \frac{3(-1) – 2(0)}{-1 + 6(-1)(0)} \)
\( \frac{dy}{dx} = \frac{-3}{-1} = 3 \)
For \(y = 3 \sin t + 2 \cos t\):
\( \frac{dy}{dt} = 3\cos t – 2\sin t \)
Final Answer:
\( \frac{dy}{dx} = 3\ at\ t = \pi \)
Solution to Part b: Equation of the Tangent at \( t = \pi \)
Key Concepts Used:
- Equation of Tangent
- Point – Slope Formula
Step-by-Step Working:
To find the tangent line equation at \( t = \pi \), start by finding values of x and y when \( t = \pi \).
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Find Coordinates at \( t = \pi \)
\( x = \sin\pi – 3\cos^2\pi = 0 – 3(-1)^2 = -3 \)
\( y = 3\sin\pi + 2\cos\pi = 0 + 2(-1) = -2 \)
So the point coordinates are: \( (-3, -2) \)
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Use Point-Slope Form
From part (a), slope \( m = 3 \):
\( y – y_1 = m(x – x_1) \)
\( y – (-2) = 3(x – (-3)) \)
\( y + 2 = 3(x + 3) \)
\( y = 3x + 7 \)
Final Answer:
\( y = 3x + 7 \)
Solution to Part c: Equation Satisfied by Parameter t at Point Q
Key Concepts Used:
- Parametric differentiation
- Tangent line intersection
Step-by-Step Working:
To show that the tangent intersects C again when \( 9\cos^2t + 2\cos t – 7 = 0 \), substitute tangent equation in the parametric equation.
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Substitute Tangent into Parametric Equations
Set \( y = 3x + 7 \) equal to \( y(t) \):
\( 3\sin t + 2\cos t = 3(\sin t – 3\cos^2t) + 7 \)
\( 3\sin t + 2\cos t = 3\sin t – 9\cos^2t + 7 \)
\( 2\cos t = -9\cos^2t + 7 \)
\( 9\cos^2t + 2\cos t – 7 = 0 \)
Final Answer:
\( 9\cos^2t + 2\cos t – 7 = 0 \)
Solution to Part d: Exact \( y \)-Coordinate of Q
Key Concepts Used:
- Solving Quadratic Equation
Step-by-Step Working:
For this part, we need to solve the quadratic in \( \cos t \) to find \( y \).
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Solve \( 9\cos^2t + 2\cos t – 7 = 0 \)
Let \( z = \cos t \):
\( 9z^2 + 2z – 7 = 0 \)
\( \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
\( z = \frac{-2 \pm \sqrt{4 + 252}}{18} = \frac{-2 \pm 16}{18} \)
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Solutions:
\( z = \frac{14}{18} = \frac{7}{9} \quad \text{&} \quad z = -1 \)
Where \( z = -1 \) corresponds to \( t = \pi \), the original point.
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Find \( y \) for \( \cos t = \frac{7}{9} \)
From \( y = 3 \sin t + 2 \cos t \), but recall the trigonometric identity
\( \sin^2 t + \cos^2 t = 1 \)
So,
\( \sin t = \sqrt{1 – \cos^2 t} = \sqrt{1 – \left( \frac{7}{9} \right)^2} = \frac{4\sqrt{2}}{9} \)
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Now, substitute in the parametric equation of variable \( y \)
\( y = 3 \sin t + 2 \cos t \)
\( y = 3 \left( \frac{4\sqrt{2}}{9} \right) + 2 \left( \frac{7}{9} \right) \)
\( y = \frac{12\sqrt{2}}{9} + \frac{14}{9} = \frac{4\sqrt{2}}{3} + \frac{14}{9} \)
Final Answer:
\( \frac{4\sqrt{2}}{3} + \frac{14}{9} \)