1. Find the expression got the square of the distance from the point P (x,y) to the point K(-1,3).
2. Find an expression for the square of the distance of P(x,y) from the line y=0.
3. The point P(x,y) moves so that it is equidistant from K and the line y=0. Show that the locus of P is a parabola and find its vertex

1. PK = sqrt[(x+1)^2 + (y-3)^2]
square of the distance = PK^2 = (x+1)^2 + (y-3)^2

2. Logically, distance of P from y=0 is always a vertical distance as y=0 is a horizontal line. Therefore, distance of P from the line y=0:
|y-0| = |y|
Square of the distance = y^2

3. From the first two parts, you have the distance of PK and distance of P from y=0
Equating the two:
(x+1)^2 + (y-3)^2 = y^2
(x+1)^2 + y^2 - 6y + 9 = y^2
(x+1)^2 = 6y - 9
(x+1)^2 = 6(y - 3/2)

Therefore the locus of P is a parabola with vertex (-1, 3/2)

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