Elimination of parameters by parametric equation x=rθ-rsinθ y=r-rcosθ Elimination parameter θ nothing

Elimination of parameters by parametric equation x=rθ-rsinθ y=r-rcosθ Elimination parameter θ nothing

Is there any mistake in the title? (2) if we get θ = arccos (R-Y) / R and then bring in (x - R θ) 2 + (Y - R) 2 = R2, then there is no θ?

The parametric equation is transformed into the general equation x = t - (2 / T) - 1, y = t + (2 / T) - 1

X = t - (2 / T) - 1, that is, x + 1 = t - (2 / T), that is, (x + 1) 2 = T2 + 4 / T2-4
Y = t + (2 / T) - 1 means y + 1 = t + (2 / T) means (y + 1) 2 = T2 + 4 / T2 + 4
① (2) introduce: (x + 1) 2 + 4 = (y + 1) 2-4

X = 2 + 3cost y = 3sint-1 is reduced to ordinary equation

If x = 2 + 3cost, then 3cost = X-2
If y = 3sint-1, then 3sint = y + 1
Then (3cost) ^ 2 + (3sint) ^ 2 = (X-2) ^ 2 + (y + 1) ^ 2 = 3 ^ 2 = 9
That is, (X-2) ^ 2 + (y + 1) ^ 2 = 9 is the circle with radius r = 3 and dot (2, - 1)

It is known that the parameter equation of C is x = 3costy = 3sint (t is the parameter), and the tangent of C at point (0, 3) is L. if the polar coordinate system is established with the origin of rectangular coordinate as the pole and the positive half axis of X axis as the polar axis, then the polar coordinate equation of L is______ ．

The parametric equation of ∵ C is x = 3costy = 3sint (t is the parameter), which is transformed into the ordinary equation x2 + y2 = 9; the equation of tangent l of circle C at point (0, 3) is y = 3; the polar equation of ∵ L is ρ sin θ = 3. So the answer is: ρ sin θ = 3