Edexcel IAL October 2022 P4 (WMA14/01) Q1 Parametric Equations
A curve C has parametric equations
\( x = \frac{1}{t-3} \qquad y = \frac{1}{t} + 2 \qquad t \in \mathbb{R},\ t > 3 \)
Show that all points on C lie on the curve with Cartesian equation
\( y = \frac{ax-1}{bx} \)
where a and b are constants to be found.
(3)
Solution to Question: Parametric Equations
Key Concepts Used:
- Eliminating the Parameter
- Solving for \( t \)
Step-by-Step Working:
-
To form the cartesian equation, eliminate \( t \) from the equation. Let’s find the value of \( t \) and substitute it to parametric equation of y-variable.
\( x = \frac{1}{t-3} \)
\( x(t-3) = 1 \)
\( xt – 3x = 1 \)
\( xt – t = 3x \)
\( t(x-1) = 3x \)
\( t = \frac{3x}{x-1} \)
- Now, substitute in equation of y.
\( y = \frac{1}{t} + 2 \)
\( y = \frac{1}{\frac{3x}{x-1}} + 2 \)
\( y = \frac{x-1}{3x} + 2 \)
\( y = \frac{x-1 + 2(3x)}{3x} \)
\( y = \frac{x-1 + 6x}{3x} \)
\( y = \frac{7x-1}{3x} \)
\( y = \frac{1}{\frac{3x}{x-1}} + 2 \)
\( y = \frac{x-1}{3x} + 2 \)
\( y = \frac{x-1 + 2(3x)}{3x} \)
\( y = \frac{x-1 + 6x}{3x} \)
\( y = \frac{7x-1}{3x} \)
Final Answer:
\( y = \frac{7x-1}{3x} \)