Question

Find the standard form of the equation of the parabola with a focus at (0, -2) and a directrix at y = 2

Answer 1

@aum

Answer 2

If you take any point, (x,y), on the parabola, it will be equidistant from the focus and the directrix. You can use this property to find the equation of the parabola.
Find the distance between (x,y) and (0,-2).
Find the distance between (x,y) and the line y = 2
Equate them.

To avoid having to take the square root when using the distance formula you can just equate the squares of the distance.

Answer 3

how do i get a point on the parabola without the parabola equation?

Answer 4

You assume a random point on the parabola and call it (x, y)
Distance-squared between (x,y) and (0,-2): (x-0)^2 + (y- -2)^2 = x^2 + (y+2)^2
Distance-squared between (x,y) and y=2: (y-2)^2
Equate them:
x^2 + (y+2)^2 = (y-2)^2
x^2 = (y-2)^2 - (y+2)^2 = (y-2+y+2)(y-2-y-2) = (2y)(-4) = -8y
y = -1/8 x^2

Answer 5

oh wow! that actually wasn't hard, thank you!

Answer 6

You are welcome.

Answer 7

do you think you could help me with another one?

Answer 8

go ahead. There is another person who has asked for help and so I will be right back.

Answer 9

alright thanks!

Answer 10

Find the vertex, focus, directrix, and focal width of the parabola.
(-1/40)x^2=y

Answer 11

If you compare it to the vertex form of a parabola: y = a(x-h)^2 + k where (h,k) is the vertex we can see that in y = (-1/40)x^2, h = 0 and k = 0.
Therefore, (0,0) is the vertex.

Answer 12

Also, comparing the given equation y = (-1/40)x^2 to y = a(x-h)^2 + k,
a = -1/40.
Find 1 / (4a).
1/(4a) = 1 / (4 * -1/40) = 1/4 * -40/1 = -10.

The focus will be 10 units below the vertex at (0, -10)

The directrix will be 10 units above the vertex at y = 10.

Answer 13

i found the focus and the directrix but idk how to find the focal width

Answer 14

Focus is at (0, -10). For the focal width, set y = -10 and find the x:
-10 = -1/40 x^2
400 = x^2
x = +20 or -20
Focal width = 20 - (-20) = 40.