A curve C has parametric equations
$x = 2t - 1,$
$y = 4t - 7 + \frac{3}{t}, \; t \neq 0$
Show that the Cartesian equation of the curve C can be written in the form
$y = \frac{2x^2 + ax + b}{x + 1}, \; x \neq -1$
where a and b are integers to be found. - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 1
Question 7
A curve C has parametric equations
$x = 2t - 1,$
$y = 4t - 7 + \frac{3}{t}, \; t \neq 0$
Show that the Cartesian equation of the curve C can be written in the form... show full transcript
Worked Solution & Example Answer:A curve C has parametric equations
$x = 2t - 1,$
$y = 4t - 7 + \frac{3}{t}, \; t \neq 0$
Show that the Cartesian equation of the curve C can be written in the form
$y = \frac{2x^2 + ax + b}{x + 1}, \; x \neq -1$
where a and b are integers to be found. - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 1
Step 1
Substituting $t = \frac{x + 1}{2}$ into $y$
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Answer
To eliminate the parameter t, we start from the equation for x:
x=2t−1⟹t=2x+1.
Now, substitute this into the expression for y:
y=4t−7+t3=4(2x+1)−7+(2x+1)3.
Simplifying this, we get:
y=2(x+1)−7+x+16=2x+2−7+x+16=2x−5+x+16.
Step 2
Writing $y$ as a single fraction
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Answer
Combine the terms to form a single fraction:
y=x+1(2x−5)(x+1)+6.
Expanding the numerator:
(2x−5)(x+1)=2x2+2x−5x−5=2x2−3x−5.
Thus, we have:
y=x+12x2−3x−5+6=x+12x2−3x+1.
Comparing this to the required format, we find:
