# Math Trig

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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 -

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 -

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 -

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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