SOLUTION: How do I find the vertex, focus, directrix, axis of symmetry, and latus rectum of this parabola equation? y - 1 = ¼(x + 2)²

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Question 421654: How do I find the vertex, focus, directrix, axis of symmetry,
and latus rectum of this parabola equation?
y - 1 = ¼(x + 2)²

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
How do I find the vertex, focus, directrix, axis of symmetry,
and latus rectum of this parabola equation?
y - 1 = ¼(x + 2)²

    y - 1 = ¼(x + 2)²

Multiply both sides by 4

4*(y - 1) = 4*¼(x + 2)²
 
4*(y - 1) = 1(x + 2)²

 4(y - 1) = (x + 2)²

Swap left and right sides:

(x + 2)² = 4(y - 1)

Compare to this standard form:

(x - h)² = 4p(y - k)

which has these properties: 

1. vertex is the point (h,k)
2. line of symmetry equation is x = h
3. focus is the point (h,k+p)
4. directrix is the line whose equation is y = k-p 
5. length of latus rectum = |4p|

h = -2, k = 1, 4p = 4, so p = 1

So for this parabola,

1. vertex is the point (h,k) = (-2,1)
2. line of symmetry equation is x = h or x = -2
3. focus is the point (h,k+p) = (-2,1+1) = (-2,2)
4. directrix is the line whose equation is y = k-p 
   or y = 1-1 or y = 0, which is the x-axis.  
5. length of latus rectum = |4p| = 4

We plot the vertex (-2,1), focus (-2,2), the line of 
symmetry (in green), and the directrix y = 0 happens 
to be the the x-axis: 



We draw the latus rectum which is a horizontal line 
segment 4p or 4 units long going through and centered 
on the focus:



Finally we draw the parabola through the ends of the 
latus rectum and through the vertex:
 


Edwin