Solve the absolute value equation algebraically, -8|-x-6|+10=2
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You can rewrite the equation in vertex form to find some of the parameters of interest.
x² +8x +4 = -4y
x² +8x +16 -12 = -4y
(x+4)² -12 = -4y
y = (-1/4)(x +4)² +3
From this, we see the vertex is (-4, 3).
The scale factor, (-1/4) is 1/(4p), where p is the distance from the vertex to the focus. Solving for p, we have
-1/4 = 1/(4p)
p = -1 . . . . . multiply by -4p
So, the focus is 1 unit below the vertex. The focus is (-4, 2).
The directrix is the same distance from the vertex, but in the opposite direction, hence the directrix is y = 4.
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On a graph, if all you know is the vertex, you can draw a line with slope 1/2 through the vertex. It intersects the parabola at the y-value of the focus. Then the distance from that point of intersection to the axis of symmetry (where the focus lies) is the same as the distance from that point to the directrix.
Every point on the parabola, including the vertex and the point of intersection just described, is the same distance from the focus and the directrix. This point of intersection is on a horizontal line through the focus, so finding the distance is made easy.
x² +8x +4 = -4y
x² +8x +16 -12 = -4y
(x+4)² -12 = -4y
y = (-1/4)(x +4)² +3
From this, we see the vertex is (-4, 3).
The scale factor, (-1/4) is 1/(4p), where p is the distance from the vertex to the focus. Solving for p, we have
-1/4 = 1/(4p)
p = -1 . . . . . multiply by -4p
So, the focus is 1 unit below the vertex. The focus is (-4, 2).
The directrix is the same distance from the vertex, but in the opposite direction, hence the directrix is y = 4.
_____
On a graph, if all you know is the vertex, you can draw a line with slope 1/2 through the vertex. It intersects the parabola at the y-value of the focus. Then the distance from that point of intersection to the axis of symmetry (where the focus lies) is the same as the distance from that point to the directrix.
Every point on the parabola, including the vertex and the point of intersection just described, is the same distance from the focus and the directrix. This point of intersection is on a horizontal line through the focus, so finding the distance is made easy.