Edexcel IAL January 2022 P4 (WMA14/01) Q3 Parametric Equations & Partial Fractions
The curve C has parametric equations
\( x = 3 + 2\sin t \qquad y = \frac{6}{7 + \cos 2t} \qquad -\frac{\pi}{2} \leq t \leq \frac{\pi}{2} \)
(a) Show that C has Cartesian equation
\( y = \frac{12}{(7-x)(1+x)} \qquad p \leq x \leq q \)
where p and q are constants to be found.
(6)
(6)
(b) Hence, find a Cartesian equation for C in the form
\( y = \frac{a}{x+b} + \frac{c}{x+d} \qquad p \leq x \leq q \)
where a, b, c and d are constants.
(3)
(3)
Solution to Part (a): Forming Cartesian Equation
Key Concepts Used:
- Trigonometric Identity: Use cos2t = 1 – 2sin²t to eliminate t.
- Substitution: Express sin t from the x-equation and substitute into the y-equation.
- Domain Consideration: Determine x-range from the given t-range.
Step-by-Step Working:
To form the cartesian equation, eliminate t from the equation.
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Apply double-angle identity
First, use the double angle formula cos 2t = 1 – 2sin²t, then substitute it in the parametric equation of variable y.
\( \cos 2t = 1 – 2\sin^2 t \)
\( y = \frac{6}{7 + \cos 2t} \)
\( y = \frac{6}{7 + (1 – 2\sin^2 t)} \)
\( y = \frac{6}{8 – 2\sin^2 t} \)
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Express sin t in terms of \( x \)
Where sin t can be found from the parametric equation of curve of x-variable.
\( x = 3 + 2\sin t \)
\( x – 3 = 2\sin t \)
\( \sin t = \frac{x-3}{2} \)
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Substitute into y-equation
\( y = \frac{6}{8 – 2\left(\frac{x-3}{2}\right)^2} \)
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Simplifying the denominator.
\( y = \frac{6}{8 – 2\left(\frac{(x-3)^2}{4}\right)} \)
\( y = \frac{6}{8 – \frac{2(x^2 – 6x + 9)}{4}} \)
\( y = \frac{6}{8 – \frac{x^2 – 6x + 9}{2}} \)
\( y = \frac{6}{\frac{16 – (x^2 – 6x + 9)}{2}} \)
\( y = \frac{12}{16 – (x^2 – 6x + 9)} \)
\( y = \frac{12}{16 – x^2 + 6x – 9} \)
\( y = \frac{12}{-x^2 + 6x + 7} \)
\( y = \frac{12}{-x^2 + 7x – x + 7} \)
\( y = \frac{12}{-x(x-7) – 1(x-7)} \)
\( y = \frac{12}{(x-7)(-x-1)} \)
\( y = \frac{12}{-1(x-7)(x+1)} \)
\( y = \frac{12}{(-x+7)(x+1)} \)
\( y = \frac{12}{(7-x)(1+x)} \)
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To determine p and q, put the values of upper and lower bound of \( t \) given in the question in x-equation to find the domain of cartesian equation. Remember, to find the domain of cartesian equation take help of the x-equation and to find range of cartesian equation use y-equation. In this part, we need domain (p and q).
For \( t = -\frac{\pi}{2} \): \( x = 3 + 2(-1) = 1 \)
For \( t = \frac{\pi}{2} \): \( x = 3 + 2(1) = 5 \)Also, the denominator \( (7-x)(1+x) \) is valid for \( x \in (1,5) \).
Hence, the domain of cartesian equation is\( 1 \leq x \leq 5 \)
Final Answer:
\( y = \frac{12}{(7-x)(1+x)} \) Shown
\( 1 \leq x \leq 5 \)
\( 1 \leq x \leq 5 \)
Solution to Part (b): Partial Fractions
Key Concepts Used:
- Trigonometric Identity: Use cos2t = 1 – 2sin²t to eliminate t.
- Partial Fraction Form: Assume \( \frac{12}{(7-x)(1+x)} = \frac{A}{7-x} + \frac{B}{1+x} \)
- Solving for Constants: Use substitution or equating coefficients method.
Step-by-Step Working:
To form the cartesian equation, eliminate t from the equation.
-
Set \( \frac{12}{(7-x)(1+x)} \) identical to \( \frac{A}{7-x} + \frac{B}{1+x} \).
\( \frac{12}{(7-x)(1+x)} = \frac{A}{7-x} + \frac{B}{1+x} \)
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Add the two fractions.
\( \frac{12}{(7-x)(1+x)} = \frac{A(1+x) + B(7-x)}{(7-x)(1+x)} \)
\( 12 = A(1+x) + B(7-x) \) -
To find B, substitute \( x = -1 \).
\( 12 = A(1-1) + B(7-(-1)) \)
\( 12 = 8B \)
\( B = \frac{3}{2} \) -
To find A, substitute \( x = 7 \).
\( 12 = A(1+7) + B(7-7) \)
\( 12 = A(8) \)
\( A = \frac{12}{8} \)
\( A = \frac{3}{2} \)
\( \frac{12}{(7-x)(1+x)} = \frac{3}{2(7-x)} + \frac{3}{2(1+x)} \)
Final Answer:
\( \frac{12}{(7-x)(1+x)} = \frac{3}{2(7-x)} + \frac{3}{2(1+x)} \)