Find the equation for the set of all points p(x,y) if it is equidistant from (2,3) and the x-axis.
Your sentence is the definition of a parabola with
focus = (2,3)
directrix y=0
Now recall that the parabola
x^2 = 4py has
vertex (0,0)
focus (0,p)
directrix y = -p
So, your parabola has
vertex at (2,3/2) with p = 3/2
So, its equation is
(x-2)^2 = 6(y-3/2)
See the graph at
https://www.wolframalpha.com/input/?i=parabola+(x-2)%5E2+%3D+6(y-3%2F2)
To find the equation for the set of all points equidistant from (2,3) and the x-axis, we can use the distance formula.
The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Since we want to find the set of points equidistant from (2,3) and the x-axis, we can write the distance formula as:
d1 = sqrt((x - 2)^2 + (y - 3)^2) (distance from (x, y) to (2,3))
d2 = sqrt((x - x)^2 + (y - 0)^2) = sqrt(x^2 + y^2) (distance from (x, y) to the x-axis)
Since the points are equidistant, we can equate d1 and d2:
sqrt((x - 2)^2 + (y - 3)^2) = sqrt(x^2 + y^2)
To eliminate the square root, we can square both sides of the equation:
(x - 2)^2 + (y - 3)^2 = x^2 + y^2
Expanding the equation:
x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 + y^2
Simplifying the equation:
-4x - 6y + 13 = 0
Therefore, the equation for the set of all points p(x,y) equidistant from (2,3) and the x-axis is -4x - 6y + 13 = 0.