AQA C4 Parametric vs. cartesian

Firstly, sorry for all the questions recently!
I'm stuck (again) on this question from June 08.

" A curve is defined by the parametric equations

x = 4t + 3

and

y = \dfrac{1}{2t} \ -1

where t does not equal 0. Find a Cartesian equation of the curve"

Now, I always thought that you could rearrange both parametric equations so that they are both "t = something" and then just make them equal to each other. However, this got me the answer

\dfrac{x-3}{4} \ = 2(y+1)

which I could then simplify to

x = 8y +11

.

This was totally different to the mark scheme which only wanted me to rearrange one of the parametric equations and came out with

(x-3)(y+1) =2

as an answer. I can't see how they got there, nor why my method is incorrect - can anyone give me a hand with this?

Thank you very much! :smile:

Reply 1

mikelbird

8

Original post by musicmedic

Firstly, sorry for all the questions recently!
I'm stuck (again) on this question from June 08.

" A curve is defined by the parametric equations

x = 4t + 3

and

y = \dfrac{1}{2t} \ -1

where t does not equal 0. Find a Cartesian equation of the curve"

Now, I always thought that you could rearrange both parametric equations so that they are both "t = something" and then just make them equal to each other. However, this got me the answer