Given the general form of a parabola x2+4x-20y+6=0, what is the distance the focus is from the vertex? p=_____.

Answer 1

20y = x^2 +4x +6

y= (1/20)x^2 + x/5 + 3/10

y = .05x^2 +.2x +.3

y= .05(x^2 +4x ) +.3

y= .05(x^2 +4x + 4) + .3 -4(.05)

y = .05(x+2)^2 + .3-.2

y = .05(x+2)^2 + .1

is the parabola's equation in vertex form: y=a(x-h)^2 +k where (h,k) is the vertex

vertex is (h,k) = (-2, 0.1)

focus is same distance from the vertex as the directrix,

vertically distance from vertex to focus = distance from vertex to directrix

as it's an upward opening parabola, but opens very slowly, it's very flat, or almost horizontal given a=.05= 1/20

a point on the parabola is (0,.3) the y intercept

from the point (0, .3) to the Focus (-2,5.1) = 5.2

from (0,.3) to the Directrix y=-4.9 is .3+4.9 = 5.2

distance from the Focus (-2, 5.1) to the Vertex (-2,.1) = 5

1/20 = 1/4p

1/5 =1/p

p=5 = 5.1-.1