It is known that the parametric equation of C is x = 3cost y = 3sint (t is the parameter), and the tangent of C at point (0,3) is o If the origin of angular coordinate is the pole and the positive half axis of X axis is the polar axis, the polar coordinate equation of O is?

It is known that the parametric equation of C is x = 3cost y = 3sint (t is the parameter), and the tangent of C at point (0,3) is o If the origin of angular coordinate is the pole and the positive half axis of X axis is the polar axis, the polar coordinate equation of O is?

Slope = 3cost / (- 3sint) | (t = π / 2) = 0
therefore
The tangent equation is
y-3=0(x-0)
Namely
y=3

In the polar coordinate system, we know that the line L passes through point a (1,0), and the minimum positive angle between its upward direction and the positive direction of the polar axis is π 3. We can find: (1) the polar coordinate equation of the line; (2) the distance from the pole to the line

(1) According to the sine theorem, it is obtained that ρ sin 2 π 3 = 1sin (π 3 − θ), that is, ρ sin (π 3 - θ) = sin 2 π 3 = 32, the polar coordinate equation of the line is ρ sin (π 3 - θ) = 32. (2) as oh ⊥ L, the perpendicular foot is h, in △ OHA, OA = 1, ∠ OHA = π 2, ∠ OAH = π 3, then Oh = oasin π 3 = 32, that is, the distance from the pole to the line is equal to 32

How to change the standard equation of P = 2cosa

This is a circle, and the equation is (x-1) ^ 2 + y ^ 2 = 1
x=p cosA
y=p sinA
x^2+y^2 = 2 x
(x-1)^2 + y^2 = 1

A problem of polar coordinate equation
The equation r = sin (2 θ - π / 2) represents a polar coordinate curve, then the range of polar angle θ is:
A [0,2π)
B [π/4,9π/4]
C [0,π/2] ∪ [π,3π/2]
D [π/4,3π/4] ∪ [5π/4,7π/4]

r=sin(2θ-π/2)=-cos(2θ)≥0
∴cos(2θ)≤0
∴2θ∈[π/2,3π/2]∪ [5π/2, 7π/2]
∴θ∈[π/4, 3π/4] ∪ [5π/4, 7π/4]
Choose D