Solving the right focus polar coordinate equation of mathematical conic Is it the same

Solving the right focus polar coordinate equation of mathematical conic Is it the same

Taking the focus F as the pole o, passing through the pole as the vertical line of the collimator L, intersecting with L at h, taking the reverse extension line ox of Oh as the polar axis, the polar coordinate system is established
Let the distance between the focus F and the guide line l be p, P (ρ, θ) be any point on the conic, connect OP and make PD ⊥ L, PQ ⊥ ox, then OP = ρ, ∠ XOP = θ
According to the definition of conic curve, there are │ op │ / │ PD │ = e, and │ PD │ = │ HQ │ = │ Ho │ + │ OQ │ = P + ρ & # 8226; Cos θ
So ρ / (P + ρ &; Cos θ) = E. that is, ρ = EP / (1-e &; Cos θ)

In the polar coordinate system, the polar coordinate equation of the line passing through the center of the circle P = 6cos @ and perpendicular to the polar axis is

p^2=6pcos@
x^2+y^2=6x
x^2-6x+y^2=0
x^2-6x+9-9+y^2=0
(x-3)^2+y^2=9
So the center of the circle (3,0)
The line is perpendicular to the polar axis
So the straight line is: x = 3
Polar equation: PCOS @ = 3
For the sake of my being so serious, just give me some points~