Find an equation for the set of points in an xy-plane that are equidistant from the point P and the line l. P(−9, 2); l: x = −3

Question

Find an equation for the set of points in an xy-plane that are equidistant from the point P and the line l.

P(−9, 2);    l: x = −3

Answer 1

Here is link to Desmos graph referenced in the video:

desmos.com/calculator/lomtybeco2

Answer 2

Points equidistant from a point and line make a parabola. The line is the directrix and the point is the focus.

As the focus is at (-9, 2) and the directrix is at x = -3, the y-coordinate of the vertex is 2 and the x co-ordinate is (-9 + -3)/2 = -6 - in between focus and directrix. Vertex is (-6, 2)

The focus is p to the right and the directrix is p to the left (which when negative means -p to the left) means

2p = -9 - -3 = -6

p = -3

Then, with p = -3 and vertex at (-6, 2), the parabola is (y - 2)^2 = 4(-3)(x - -6)

(y-2)^2 = -12(x+6)

If you want to write this in standard form, you expand the 2 sides.

y^2 - 4y +4 = -12x -72

y^2 -4y+76 = -12x

x = -1/12y^2 + 1/3y - 6 1/3

As a check of your match, substitute y = 2, and you get x = -6.