Find the vertex, focus, and the directrix of the parabola

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SUMMARY

This discussion focuses on finding the vertex, focus, and directrix of parabolas represented by equations such as \(y^2 = -6x\) and \(x + y^2 = 0\). The method involves completing the square and transforming the equations as necessary. For the equation \(y^2 = -6x\), the vertex is at (0,0), the focus is at (0, -3/2), and the directrix is the line \(x = -3/2\). The discussion emphasizes reversing x and y in certain equations to simplify the process of finding these parameters.

PREREQUISITES
  • Understanding of parabolic equations and their standard forms
  • Knowledge of completing the square technique
  • Familiarity with the concepts of vertex, focus, and directrix
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the standard form of parabolic equations: \( (x-h)^2 = 4p(y-k) \)
  • Learn how to complete the square for quadratic equations
  • Explore transformations of conic sections, particularly switching x and y
  • Research the properties of parabolas, including directrix and focus relationships
USEFUL FOR

Students of algebra, mathematics educators, and anyone interested in mastering the properties and characteristics of parabolas in coordinate geometry.

konartist
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Find the vertex, focus, and the directrix of the parabola.

I get tripped up sometimes, but I know how to find all of the stuff with an equation like this :
(x+1)^2 +8(y+3) = 0

But how do I find it with equations like this
y^2=-6x

or

x+y^2=0

Can you complete the square with only two terms?
 
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konartist said:
Find the vertex, focus, and the directrix of the parabola.

I get tripped up sometimes, but I know how to find all of the stuff with an equation like this :
(x+1)^2 +8(y+3) = 0

But how do I find it with equations like this
y^2=-6x

or

x+y^2=0

Can you complete the square with only two terms?

Reverse x and y! That is, change x+ y2= 0 to x2+ y= 0 which you know how to do. Once you have found the vertex, focus, directrix for that, switch back: If you found (a,b) as the vertex of x2+y= 0, then (b,a) is the vertex of x+ y2= 0. If you found y= c as the directrix of x2+ y= 0, then the directrix of
y2+ x= 0 is x= c.
 
My work

(x-h)^2=4p(y-k)
since h and k are 0 then the vertex = (0,0)
How do I find the focus and the directrix though?
Focus maybe = 4p = -6 P= -3/2 so the focus (0,-3/2) ?
Directrix ?
 

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