Parabola conical whatever equations

  • Thread starter Thread starter runicle
  • Start date Start date
  • Tags Tags
    Conical Parabola
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 3K views
runicle
Messages
75
Reaction score
0
I am dazed and confused how does 2x^2=y end up being directrix y=-1/8?
 
Physics news on Phys.org
I forgot the steps on how to do this.
 
runicle said:
I am dazed and confused how does 2x^2=y end up being directrix y=-1/8?
Putting the parabola into form [itex]x^2 = 4py[/itex] where p is the focus, will give you the location of the focus, p. Since the points on the parabola are equidistant form the directrix and focus, that should enable you to find the equation of that line.

AM
 
I don't see the math in it though...
[itex]x^2 = 4py[/itex]
[itex]x/4^2 = py[/itex]
making [itex]x/4^2 = p[/itex]
is it something like that?
 
runicle said:
I don't see the math in it though...
[itex]x^2 = 4py[/itex]
[itex]x/4^2 = py[/itex]
making [itex]x/4^2 = p[/itex]
is it something like that?
No. [itex]2x^2 = y[/itex] so [itex]x^2 = \frac{1}{2}y[/itex]. In the form [itex]x^2 = 4py[/itex] 4p = 1/2 and p = 1/8. So the focus is (0,1/8). Since the vertex is (0,0) which is equidistant from the focus and the directrix, the directrix is the horizontal line passing through (0,-1/8) ie. y = -1/8

AM