Convert the polar equation... r=1/1-sin2θ to rectangular form?

0 ⤊

im really stuck. i tried to what i thought was everything. please help

2007-10-24 13:59:01 · 3 answers · asked by Andrew T 2 in Science & Mathematics ➔ Mathematics

Hint
x = r cosθ , y = r sinθ
( r cosθ , r sinθ ) ð ( x, y )
Hope that helps

2007-10-24 14:04:10 · answer #1 · answered by Minase D 2 · 0⤊ 0⤋

r= 1/(1-2x*y/r^2) so 1= r/(r^2-2xy) remember sin 2t = 2 sint*cost

x^2+y^2-2xy =sqrt(x^2+y^2)
(x-y)^4 =x^2+y^2

2007-10-24 14:12:59 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋

r = sqrt(x^2 + y^2)
sin(2u) = 2sin(u)cos(u) = 2xy/(x^2 + y^2)

sqrt(x^2+y^2)= 1/(1 - 2xy/(x^2+y^2))

2007-10-24 14:07:06 · answer #3 · answered by Anonymous · 0⤊ 0⤋