focus at (-2,0) directrix the line x=2
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eg
https://www.khanacademy.org/math/algebra2/conics_precalc/parabolas_precalc/v/using-the-focus-and-directrix-to-find-the-equation-of-a-parabola
To find the equation of the parabola with a focus at (-2,0) and a directrix at x=2, we can start by understanding the definition of a parabola.
A parabola is a set of all points that are equidistant from the focus and the directrix.
For this problem, the focus is at (-2,0) and the directrix is the line x=2. The directrix is a vertical line since it has a constant x-coordinate.
To find the equation of the parabola, we first need to find the distance between a point (x, y) on the parabola and the focus (-2,0). We'll call this distance "d".
Using the distance formula, the distance d between a point (x, y) and the focus (-2,0) is given by:
d = sqrt((x - (-2))^2 + (y - 0)^2)
= sqrt((x + 2)^2 + y^2)
We also need to find the distance between a point (x, y) on the parabola and the directrix x = 2. We'll call this distance "d'".
Since the directrix is a vertical line with x = 2, the distance d' between a point (x, y) and the line x = 2 is equal to the difference between the x-coordinate of the point and the x-coordinate of any point on the directrix. In this case, we can use the x-coordinate of the point on the directrix as 2. So,
d' = |x - 2|
Now, according to the definition of a parabola, these distances should be equal. Therefore, we have:
sqrt((x + 2)^2 + y^2) = |x - 2|
Squaring both sides of the equation, we get:
(x + 2)^2 + y^2 = (x - 2)^2
Expanding and simplifying, we have:
x^2 + 4x + 4 + y^2 = x^2 - 4x + 4
Simplifying further, the x^2 terms cancel out, giving:
4x + 4 + y^2 = -4x + 4
Rearranging the terms, we have:
8x + y^2 = 0
So, the equation of the parabola with a focus at (-2,0) and a directrix at x=2 is 8x + y^2 = 0.