Find equation of curve in cartesian coordnts describing all points equidistant from (1,0) + y=-2?

Question

Find equation of curve in cartesian coordinates describing all points in the plane which are equidistant from (1,0) and y= -2

Answer 1

this is a parabola...
focus: (1,0) ... directrix: y = -2

the vertex is at (1, -1)
p = 1

4(y+1) = (x-1)^2

§

Answer 2

Suppose (x,y) is on the curve.

The distance between (x,y) and (1,0) is sqrt((x-1)² + (y-0)²)

The distance between (x,y) and the line y = -2 is | y - (-2) |

These distances are to be equal.

After simplifying a bit, we have

| y + 2 | = sqrt((x-1)² + y²)

Square both sides:

(y + 2)² = (x - 1)² + y²

This is a parabola. If you want the "vertex" form of the equation, expand the (y + 2)² term but not the (x-1)² term. Otherwise, expand both terms and simplify.