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C4 - Cartesian/Parametric Q help please

How would you convert x=t^3 - 8t and y=t^2 into its cartesian form?

I was thinking of sq rotting the y to make root y = 2 but then it could be + or minus?

Thanks :smile:

Try squaring the x.

Surely square rooting y would give sqrt(y)=t? then substitute that back into the x formula and replace all t's with sqrt(y)

Reply 3

Cleoleo

Original post by lamalas600

Surely square rooting y would give sqrt(y)=t? then substitute that back into the x formula and replace all t's with sqrt(y)

but could it not be +/- sq root y?

Reply 4

Hazza616

You don't have to worry about that, as you will be squaring the Y again later on.

Reply 5

Mathlete 4 the win

squaring the x and then subbing in y would work

Reply 6

Cleoleo

Original post by 4 Mathlete the win

squaring the x and then subbing in y would work

So squaring the whole of x=t^3 - 8t to get t^6 - 64t^2 = x^2 and then subbing y where you see t^2... so it would be x^2 = y^3 - 64y ?

Reply 7

ghostwalker

Original post by Cleoleo

So squaring the whole of x=t^3 - 8t to get t^6 - 64t^2 = x^2

Have another go:

x^2=(t^3-8t)^2=(t^3-8t)\times(t^3-8t)=....

Reply 8

Mathlete 4 the win

Original post by Cleoleo

So squaring the whole of x=t^3 - 8t to get t^6 - 64t^2 = x^2 and then subbing y where you see t^2... so it would be x^2 = y^3 - 64y ?

Yeah as ghostwalker has said you haven't quite expanded it correctly but after that you have the right idea. Just expand it as you would normally making sure not miss any terms.

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