Parabola conical whatever equations

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    Conical Parabola
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The discussion centers on finding the directrix of the parabola defined by the equation 2x^2 = y. To convert this into the standard form x^2 = 4py, it is established that 4p equals 1/2, leading to p being 1/8. Consequently, the focus of the parabola is located at (0, 1/8). Since the vertex is at (0, 0), the directrix is determined to be the line y = -1/8. The mathematical steps clarify the relationship between the focus, vertex, and directrix of the parabola.
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I am dazed and confused how does 2x^2=y end up being directrix y=-1/8?
 
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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 x^2 = 4py 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...
x^2 = 4py
x/4^2 = py
making x/4^2 = p
is it something like that?
 
runicle said:
I don't see the math in it though...
x^2 = 4py
x/4^2 = py
making x/4^2 = p
is it something like that?
No. 2x^2 = y so x^2 = \frac{1}{2}y. In the form x^2 = 4py 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
 

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