SOLUTION: what is the vertex directrix and focus of the equation (x+5)^2=4y

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Question 1012164: what is the vertex directrix and focus of the equation (x+5)^2=4y
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%2B5%29%5E2=4y+
y=%281%2F4%29%28x%2B5%29%5E2+....compare to y=%28x-h%29%5E2%2Bk and you see that h=-5 and k=0
so, vertex is at (-5,0
or, we can use standard form and calculate coordinates of the vertex like this:
y=%281%2F4%29%28x%5E2%2B10x%2B25%29+
y=%281%2F4%29x%5E2%2B%281%2F4%2910x%2B%281%2F4%2925+
y=%281%2F4%29x%5E2%2B%285%2F2%29x%2B%2825%2F4%29+
the x-coordinate of the focus is:
x+=+-b+%2F%282a%29
x+=+-%285%2F2%29%2F+%282%281%2F4%29%29
x+=+-%285%2F2%29%2F+%281%2F2%29
x+=+-%2810%2F2%29
x+=+-5
then, y-coordinate is
y=%281%2F4%29%28-5%29%5E2%2B%285%2F2%29%28-5%29%2B%2825%2F4%29+
y=%281%2F4%2925-%2825%2F2%29%2B%2825%2F4%29+
y=25%2F4-50%2F4%2B%2825%2F4%29+
y=50%2F4-50%2F4+
y=0+
so, we got same result for vertex

the focus is:
The focus lies on the axis of symmetry of the parabola, and has the y-coordinate k%2B1%2F%284a%29. Because we just found the vertex to be (-5,0), we know the axis of symmetry to be x=-5, and the focus lies on that line.
y-coordinate is at 1%2F%284a%29+=1%2F%284%281%2F4%29%29=1%2F%284%2F4%29=1
and the coordinates of the focus are:(-5,1)

Once you know the y=coordinate of the vertex, k=0, it is given by y+=+k+%96+p, where p+=+1%2F%284a%29. Thus, as we calculated for the focus, above:
p+=+1%2F%284a%29+=+1%2F%284%2A%281%2F4%29%29+=+1
and the directrix is y+=+0+%96+1+=-1