Page 4c

EXTENSIONS Menu

MAIN MENU

Historically, the locus definition was used to find the standard equation of a parabola.

Let's look at a few examples to see how this was done.

EXAMPLE 1:
The parabola in Figure 1 has its vertex at (3, -1), its focal point at (3, 2) and a directrix at y = -4. Find the standard equation of this parabola using the locus definition.	Figure 1
In order to use the locus definition PF = PD to find the standard equation of this parabola, you must first find some important points and distances.
Using the distance formula, you can calculate the length of PF using P = (x, y) and the focal point F = (3, 2). This gives: PF = [(x - 3)2 + (y - 2)2]½.	Figure 2
In order to find the length of PD, you must first find point D on the directrix by drawing a perpendicular line from P(x, y) to the directrix. The point where the perpendicular line crosses the directrix is point D. Since the equation of the directrix is y = -4, the coordinates of D are (x, -4).
Using the distance formula, you can now find the length of PD using the points P = P(x, y) and D = (x, -4). This gives: PD = [(x - x)2 + (y + 4)2]½ = [(y + 4)2]½.
Therefore, Let PF = PD [(x - 3)2 + (y - 2)2]½ = [(y + 4)2]½ Square both sides: (x - 3)2 + (y - 2)2 = (y + 4)2 Expand: x2 - 6x + 9 + y2 - 4y + 4 = y2 + 8y + 16 Simplify and collect like terms:   x2 - 6x = 12y + 3 Complete the square: x2 - 6x + 9 = 12y + 3 + 9 Simplify: (x - 3)2 = 12y + 12 Put in standard form: y = (1/12)(x - 3)2 - 1
So the standard equation of the above parabola is y = (1/12)(x - 3)2 - 1.
Look back at Figure 1. Notice that the distance between the focal point and the vertex is 3 units. This is also the distance between the vertex and the directrix.
What do you notice about this distance and the a = 1/12 in the standard equation y = (1/12)(x - 3)2 - 1?
It's tricky, but the important thing to notice is that 12 = 4 × 3, where 3, remember, is the distance from the focus to the vertex. So a =  1 =    1    12 4 × 3

Summary

Notice that in all the examples and questions: 4 × [half the distance between the focal point and the directrix] = 1/a where a is the same a as in the standard equation y - k = a(x - h)2 or x - h = a(y - k)2

If we let p = half the distance between the focal point and the directrix, then we can replace 1/a with 4p in the standard equation. Thus the standard equation becomes (x - h)2 = 4p(y - k) or (y - k)2 = 4p(x - h).

Note that if a parabola's vertex is located at the origin, this standard equation is x2 = 4py or y2 = 4px.