Eliminate the parameter to find a Cartesian equation for #x = sin^2 t# and #y=2cost#?

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Answer 1 · 2017-10-01T05:13:20

#y^2 = -4(x-1)#

There are many ways to do this. I'll choose one which relies upon using a common trigonometric identity, namely:

# sin^2 t + cos^2 t = 1 #

We already have an expression for #sin^2 t# - namely, #x# - so all that remains is to manipulate the other expression to fit that pattern:

#y = 2cos t#

#y/2 = cos t #

#cos^2 t = y^2/4#

Take these two relationships and place them into the trig identity:

# sin^2 t + cos^2 t = 1 #

# x + y^2/4 = 1 #

# 4x + y^2 = 4 #

#y^2 = -4x + 4# or #y^2 = -4(x-1)#

(By the way, this indicates the result is a parabola which opens to the left, with a vertex at #(1,0)#