Math Trig
posted by kyu .
Find the Cartesian form of the parametric equation.
x = (2a)(cot T)
y = (2a)(sin^2 T)
how? lol
here's what i got
y = (sin^2 T)
y = (2a)y^2
y = 2a
then?
do the same for X?
i'm stuck there

Proceed to eliminate T from the two equations, you will end up with a single equation involving x and y. Solve for y.
x=(2a)cot(T)....(1a)
x² = (4a²)cot²(T)...(1b)
Using cot²(x)+1 = csc²(x)
we get cot²(x)=csc²(x)1
1(b) becomes
x² = (4a²)(csc²(T)1)
or
sin²(T) = 4a²/(4a²+x²).....(1c)
From
y = (2a)(sin^2 T)
we get
sin²(T) = y/(2a) ....(2a)
Substitute (2a) in (1c)
y/(2a) = 4a²/(4a²+x²)
y=8a³/(4a²+x²) 
how'd you get (4a²+x²) from
x² = (4a²)(csc²(T)1) ?
i know you can x² = sin²(T) but
how'd you get
(4a²+x²)?? 
By moving the "1" term to the left hand side, we end up with only one term containing T:
x² = (4a²)(csc²(T)1)
x² = 4a²/sin²(T)  4a²
x²+4a² = 4a²/sin²(T)
sin²(T) = 4a²/(x²+4a²)