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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(20064)
      (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
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