(1)y2=4x;(2)y2+x2-2x-1=0;(3)θ=
;(4)ρcos2
=1;(5)ρ2cos2θ=4;(6)ρ=
.
(1)y2=4x;(2)y2+x2-2x-1=0;(3)θ=;(4)ρcos2=1;(5)ρ2cos2θ=4;(6)ρ=.
解:(1)将x=ρcosθ,y=ρsinθ代入y2=4x,得(ρsinθ)2=4ρcosθ,
化简得ρsin2θ=4cosθ.
(2)将x=ρcosθ,y=ρsinθ代入y2+x2-2x-1=0,得(ρsinθ)2+(ρcosθ)2-2ρcosθ-1=0,
化简得ρ2-2ρcosθ-1=0.
(3)tanθ=,∴tan==
,化简得y=
x(x≥0).
(4)∵ρcos2
=1,
∴ρ
=1,即ρ+ρcosθ=2.
∴
+x=2,化简得y2=-4(x-1).
(5)∵ρ2cos2θ=4,∴ρ2cos2θ-ρ2sin2θ=4,即x2-y2=4.
(6)∵ρ=
,
∴2ρ-ρcosθ=1.
∴2
,化简得3x2+4y2-2x-1=0.