Posted by **kyu** on Monday, May 31, 2010 at 5:44pm.

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

- Math Trig -
**MathMate**, Monday, May 31, 2010 at 7:29pm
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)

- Math Trig -
**kyu**, Tuesday, June 1, 2010 at 12:27am
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)??

- Math Trig -
**MathMate**, Tuesday, June 1, 2010 at 7:39am
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)

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