Parabola conical whatever equations

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    Conical Parabola
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SUMMARY

The discussion centers on deriving the directrix of the parabola represented by the equation 2x² = y. By converting this equation into the standard form x² = 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), and the directrix is the horizontal line y = -1/8. This process highlights the relationship between the vertex, focus, and directrix of a parabola.

PREREQUISITES
  • Understanding of quadratic equations and their standard forms
  • Knowledge of parabolic geometry, specifically focus and directrix concepts
  • Familiarity with algebraic manipulation of equations
  • Basic skills in graphing parabolas
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  • Study the derivation of the standard form of a parabola from its general equation
  • Learn about the properties of parabolas, including focus, directrix, and vertex
  • Explore applications of parabolas in physics and engineering contexts
  • Practice solving various parabola-related problems using different equations
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Students studying algebra and geometry, educators teaching quadratic functions, and anyone interested in the mathematical properties of parabolas.

runicle
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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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