Find the coordinates of the vertex and focus and the equation of the directrix for the parabola y^2+8x+40=0
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Find the coordinates of the vertex and focus and the equation of the directrix for the parabola y^2+8x+40=0

[From: ] [author: ] [Date: 12-01-09] [Hit: ]
and the directrix is the vertical line two units to the right of the vertex, x = -3.......
Question in the title, i'm very confused, please help :) <3

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y^2+8x+40=0 => sideways parabola, opens to the left:
If the parabola is rotated so that its vertex is (h,k) and its axis of symmetry is parallel to the x-axis, it has an equation of (y - k)^2 = 4p (x - h), where the focus is (h + p, k) and the directrix is x = h - p
y^2 = -8(x + 5) => vertex at(-5 , 0)
4p = -8
p = -2
x = h - p = -5 + 2 = -3 => directrix

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Rephrase the equation as follows:

-(y - 0)^2 = (4)(2)(x + 5)

This tells us that the vertex is at (-5,0), and that the focus and directrix are 2 units away. Since this is a parabola opening to the left, the focus is two units left of the vertex at (-7.0), and the directrix is the vertical line two units to the right of the vertex, x = -3.
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