SOLUTION: Write an equation for the set of all points in the plane equidistant from (-4.-13/2) and x=4.

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Question 1138876: Write an equation for the set of all points in the plane equidistant from (-4.-13/2) and x=4.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The set of points equidistant from a fixed point (focus) and fixed line (directrix) is a parabola.

With a vertical directrix at x=4 and a focus at (-4,-6.5), the parabola opens to the left.

The vertex is the point on the axis of symmetry equidistant from the focus and directrix, so the vertex is (0,-6.5).

The equation of a parabola opening left or right with vertex at (0,-6.5) is

%28x-0%29+=+%281%2F%284p%29%29%28y-%28-6.5%29%29%5E2

x+=+%281%2F%284p%29%29%28y%2B6.5%29%5E2

In the equation in that form, p is the directed distance from the vertex to the focus. With the vertex at x=0 and the focus at x=-4, p = -4.

So 4p = -16, and the completed equation is

x+=+%28-1%2F16%29%28y%2B6.5%29%5E2