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

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.

    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)
    sin²(T) = 4a²/(4a²+x²).....(1c)

    y = (2a)(sin^2 T)
    we get
    sin²(T) = y/(2a) ....(2a)

    Substitute (2a) in (1c)
    y/(2a) = 4a²/(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

  • 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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