Parabola conical whatever equations

  • Thread starter Thread starter runicle
  • Start date Start date
  • Tags Tags
    Conical Parabola
Click For Summary

Homework Help Overview

The discussion revolves around the properties and equations of parabolas, specifically focusing on the equation 2x² = y and its relation to the directrix and focus of the parabola.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants express confusion regarding the transformation of the parabola equation into standard form and the derivation of the directrix. There are attempts to clarify the relationship between the focus, directrix, and the vertex of the parabola.

Discussion Status

Some participants are exploring the mathematical relationships involved, while others are questioning the steps taken to derive the directrix from the given equation. There is an ongoing exchange of ideas, with some guidance being offered regarding the standard form of the parabola.

Contextual Notes

Participants are working under the constraints of recalling specific mathematical steps and definitions related to parabolas, which may not be fully clear to everyone involved.

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
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K