SOLUTION: P(x,y)is a variable point and A(2,2) is a fixed point. Find the equation of the locust of P if it moves so that AP is always equal to the perpendicular distance from the line y=3.I

Question 1085118: P(x,y)is a variable point and A(2,2) is a fixed point. Find the equation of the locust of P if it moves so that AP is always equal to the perpendicular distance from the line y=3.Identity the locus and show it on the diagram.
When is this locus of P less than zero?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Note: the term is locus (not locust)

Given points
A = (2,2)
P = (x,y)

Define point B as B = (x,3) where the 'x' is the same as the x coordinate of point P. This point B is directly above point P. Point B is located on the horizontal line y = 3.
For a visual, see the diagram shown at the bottom of the page.

The distance from B to P is notated as d(B,P)
The distance from A to P is notated as d(A,P)

In this case,
d(B,P) = 3-y
since this is the vertical distance from B to P
Also,
d(A,P) = sqrt((x-2)^2+(y-2)^2)
using the distance formula

Equate the two distance expressions and solve for y
d(B,P) = d(A,P)
3-y = sqrt((x-2)^2+(y-2)^2)
(3-y)^2 = (x-2)^2+(y-2)^2
9 - 6y + y^2 = x^2 - 4x + 4 + y^2 - 4y + 4
9 - 6y = x^2 - 4x + 4 - 4y + 4
-2y = x^2 - 4x - 1
y = (-1/2)*(x^2 - 4x - 1)
y = (-1/2)*(x^2) + (-1/2)*(-4x) + (-1/2)*(-1)
y = -0.5x^2 + 2x + 0.5

The equation is a parabola. The specific equation is y = -0.5x^2 + 2x + 0.5 where P is the point on the parabola (allowed to slide around) and point A is the focus (fixed). The line y = 3 is the directrix.

This is what the graph would look like.

Image generated by GeoGebra (free graphing software).

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