Question

What is the equation of a parabola with a directrix x = 8 and focus (-8, 0).

Answer 1

http://www.mathwarehouse.com/quadratic/parabola/focus-and-directrix-of-parabola.php

Answer 2

give me a second, reading up on how to do this.

Answer 3

k

Answer 4

Let's follow along this example.

Answer 5

Let (x,y) be any point on the parabola. Find the distance between (x,y) and the focus, (-8,0):
sqrt((x+8)^2+(y-0)^2) = sqrt((x+8)^2+y^2).
The distance between (x,y) and the directrix, x = 8, is:
|x-8|.
Equate the two distance expressions and square both sides:
sqrt((x+8)^2+y^2) = |x-8|
(x+8)^2 + y^2 = (x-8)^2

Answer 6

wait let me double check everything.

Answer 7

cool

Answer 8

ok, now I'm lost lol.. @myininaya @TuringTest @dpaInc does any of you remember how to do this?

Answer 9

*do

Answer 10

@Directrix

Answer 11

@jim_thompson5910

Answer 12

Can you bump this?

Answer 13

The vertex is the midpoint of the segment that is perpendicular to the directrix and that goes through the focus

So the vertex has a y coordinate of 0

The x coordinate of the vertex is (8 + (-8))/2 = 0/2 = 0

which means that the vertex is at (0,0)

Answer 14

not yet.. in 15 min :/

Answer 15

Now use the form

4p(x-h) = (y-k)^2

where p is the distance from the focus to the vertex (along the line of symmetry) and (h,k) is the vertex

Answer 16

since the directrix is a vertical line, this means the parabola will have the equation:
\(\large (y-k)^2=4a(x-h) \)
where (h,k) is your vertex....