Question

Find the standard form of the equation of the parabola with a focus at (0, -10) and a directrix at y = 10

Answer 1

@jim_thompson5910 HALP

Answer 2

how far is the point (0, -10) away from the directrix y = 10

Answer 3

20

Answer 4

cut this in half to get 20/2 = 10

Answer 5

the vertex will be 10 units away from the focus

AND

be 10 units away from the directrix

Answer 6

so the vertex will lie on the midpoint of the segment connecting the focus and the directrix (this segment is also perpendicular to the directrix)

Answer 7

so where is the vertex?

Answer 8

0,0

Answer 9

good

Answer 10

the vertex in general is (h,k)

so if the vertex is (0,0), then

(h,k) = 0

h = 0
k = 0

Answer 11

the distance from the vertex to the focus (call it p) is

p = 10

Answer 12

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

4*10(y - 0) = (x - 0)^2 ... plug in the three values just found

40y = x^2

y = (1/40)*x^2

Answer 13

oh wait, i have my signs mixed up

Answer 14

THANKS YOU'RE A GREAT MAN @jim_thompson5910

Answer 15

it should be

y = -(1/40)*x^2

because the parabola opens downward

Answer 16

If the telescope is 1 m deep and 12 m wide, how far is the focus from the vertex? ONE MORE

Answer 17

@jim_thompson5910

Answer 18

let me think

Answer 19

ok you'll have to do a drawing of a parabola

Answer 20

but you'll have a parabola going through these 3 points

(0,0)
(6,1)
(-6,1)

Answer 21

done

Answer 22

ok great

Answer 23

The point (0,0) is the vertex. We'll use (6,1) as another point

y = a(x-h)^2 + k

y = a(x-0)^2 + 0

y = ax^2

1 = a(6)^2

1 = 36a

a = 1/36

So the equation of the parabola is y = (1/36)x^2

----------------------------------------------------

Now convert it to 4p(y - k) = (x - h)^2 form


y = (1/36)x^2

y - 0 = (1/36)(x - 0)^2

36(y - 0) = (x - 0)^2

4*9(y - 0) = (x - 0)^2

It's now in 4p(y - k) = (x - h)^2 form where

h = 0
k = 0
p = 9

So the distance from the vertex to the focus is 9 meters

Answer 24

yay thanks so much!

Answer 25

np