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x=3cost y=9sin2t

Find the cartesian equation in the form y^2 =f(x)

I know you make cost subject and sint subject and then do (sint)^2 + (cost)^2 = 1 but you have sin2t

By Averagemathgeek (Averagemathgeek) on Monday, March 08, 2004 - 03:10 pm: Edit

y=9sin(2t)=9*2cos(t)sin(t)=[3cos(t)]*6sin(t)=6*x*sin(t)
x=3cos(t), t=arccos(x/3)
y=6*x*sin[arccos(x/3)]=6*x*sqrt(1-x^2/9)
y^2=36*x^2*(1-x^2/9)=x^2*(36-4x^2)=36x^2-4x^4

Hope this helps.