[Cleoleo]

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

[ghostwalker]

Try squaring the x.

[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)

[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?

[Hazza616]

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

[Mathlete 4 the win]

squaring the x and then subbing in y would work

[Cleoleo]

(Original post by Mathlete 4 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 ?

[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:

[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.