拋物線的證明
Author: 藍邦偉  老師

試證以F(c,0)為焦點,\;L:x=-c為準線的拋物線方程式為\;\Gamma：\;y²=4cx

pf:

設P(x,y)∈\Gamma

∵\overline{PF}=d(P,L)

∴\sqrt{(x-c)²+y²}=|x+c|

⇒x²-2cx+c²+y²=x²+2cx+c²

⇒y²=4cx

$ L $

A = (-3.94, 6.2)
Step = 0 (slider from 0 to 5, step 1)
O = (9.9, 1.42)
u = Vector(O - (3, 0), O + (3, 0))
v = Vector(O - (0, 3), O + (0, 3))
F: O + (1, 0)
a = Segment(O - (1, 3), O + (-1, 3))
c = Parabola(F, a)
P = Point(c)
b = Segment(P, F)
P_{H}: (x(O) - 1, y(P))
d = Segment(P_{H}, P)
e = Circle(P, F)
n = 0