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

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To find the vertex, focus, and directrix of parabolas given in different forms, one can utilize the method of completing the square. For the equation y^2 = -6x, the vertex is at (0,0), and by determining p from the equation, the focus is located at (−3/2, 0) and the directrix is x = 3/2. When working with equations like x + y^2 = 0, it's helpful to switch the variables to apply known methods. The process involves reversing x and y, finding the parameters for the transformed equation, and then reverting to the original variables for the final results. Understanding these transformations simplifies the identification of key parabola characteristics.
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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