SOLUTION: please i want you to help me find the equation of the locus of points equidistant from the point(0,2) and the straight line y=4.

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Question 947097: please i want you to help me find the equation of the locus of points equidistant from the point(0,2) and the straight line y=4.
Found 3 solutions by Theo, ikleyn, greenestamps:
Answer by Theo(13342)   (Show Source):
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if you draw a circle around the point (0,2) with a radius of 2, you will find that you have 2 points that are the same distance from the point (0,2) and also the same distance from the line y = 4.

those points are (-2,2) and (2,2)

they are 2 units away from the point (0,2).

they are 2 units away from the line y = 4.

their distance from the line y = 4 is measured by drawing a line perpendicular to the line y = 4 from each of those points.

those vertical lines will intersect the line y = 4 at the points (-2,4) and (2,4).

(-2,2) is 2 units away from (-2,4).

(2,2) is 2 units away from (2,4).

the points (-2,2) and (2,2) are therefore both the same distance from the point (0,2) and the line y = 4.

a picture of what this looks like is shown below:

Answer by ikleyn(52786)   (Show Source):
You can put this solution on YOUR website!
.

Let (x,y) be the point of the locus. Then the distance to the point (0,2) is d = = . while the distance to the straight line y= 4 is |y-4|. Two distances are the same; it gives an equation = |y-4|. Square both sides x^2 + y^2 - 4y + 4 = y^2 - 8y + 16. Simplify x^2 = - 4y + 12, or 4y = - x^2 + 12, or y = - . It is the parabola opened downward, with the symmetry axis x= 0, with the vertex at the point (0,3) and x-intercepts at x= - and x= . Plot y = - (red) and y = 4 (green)

Solved.

Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!

The given conditions are those for a parabola with directrix y=4 and focus (0,2).

With directrix at y=4 and focus at (0,2), the parabola opens downward. The vertex of the parabola is halfway between the focus and directrix, at (0,3).

The vertex form of the equation is

or

where (h,k) is the vertex and p is the directed distance (i.e., can be negative) from the directrix to the vertex, or from the vertex to the focus.

The given conditions tell us (h,k) is (0,3) and p is -1. So the equation is

or