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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

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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.png

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... 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 for t

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Answer

First, solve for t in terms of x from the equation x=2t1x = 2t - 1:

t=x+12t = \frac{x + 1}{2}

Now, substitute this expression for t into the equation for y:

y=4(x+12)7+3(x+12)y = 4\left(\frac{x + 1}{2}\right) - 7 + \frac{3}{\left(\frac{x + 1}{2}\right)}

This simplifies to:

y=2(x+1)7+6x+1y = 2(x + 1) - 7 + \frac{6}{x + 1}

Step 2

Combining terms

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Answer

Combine the terms in y:

y=2x+27+6x+1y = 2x + 2 - 7 + \frac{6}{x + 1}

Which can be simplified to:

y=2x5+6x+1y = 2x - 5 + \frac{6}{x + 1}

Now, we need to express y as a single fraction:

y=(2x5)(x+1)+6x+1y = \frac{(2x - 5)(x + 1) + 6}{x + 1}

Step 3

Final expression

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Answer

Expanding the numerator:

(2x5)(x+1)=2x2+2x5x5=2x23x5(2x - 5)(x + 1) = 2x^{2} + 2x - 5x - 5 = 2x^{2} - 3x - 5

Thus, we have:

y=2x23x5+6x+1=2x23x+1x+1y = \frac{2x^{2} - 3x - 5 + 6}{x + 1} = \frac{2x^{2} - 3x + 1}{x + 1}

Comparing this with the required form:

y=2x2+ax+bx+1y = \frac{2x^{2} + ax + b}{x + 1}

We identify that:

a = -3, b = 1.

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