Question

Find the standard form of the equation of the parabola with a focus at (-4, 0) and a directrix at x = 4.

Answer 1

@sleepyhead314

Answer 2

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

Answer 3

The vertex is always halfway between the focus and the directrix, and the parabola always curves away from the directrix, so I'll do a quick graph showing the focus, the directrix, and a rough idea of where the parabola will go:

Answer 4

yes, that is what they give me to use

Answer 5

*cheers* go alex go! :D
cuz idk how the heck to do this xD

Answer 6

(y – (–2))^2 = 4(–1)(x – 1), or (y + 2)^2 = –4(x – 1)

Answer 7

you is copy pasting a lot @AlexandervonHumboldt2 ?

Answer 8

ok l;et me explain another way

Answer 9

Any point, (x0 , y0) on the parabola satisfies the definition of parabola, so there are two distances to calculate:

Distance between the point on the parabola to the focus
Distance between the point on the parabola to the directrix

Answer 10

@AlexandervonHumboldt2 would the vertex for this be (0, 0) ?

Answer 11

Distance between the point (x0 , y0) and (a , b): where (a, b) is the focus. Use distance formula here: sqrt((x0-a)^2+(y0-b)^2)

Answer 12

no

Answer 13

Distance between point ( x0 , y0) and the line y = c : =|y0-c|

Answer 14

Evaluate: sqrt((x0-a)^2+(y0-b)^2)=|y0-c|

Answer 15

Square both sides.

Answer 16

I am very confused by all this terminology @AlexandervonHumboldt2
you are typing very very fast

Answer 17

iambatman you still havent answered my question ._.

Answer 18

*screams head off*

Answer 19

*rips hair out*

Answer 20

Expand the expression in y0 on both sides and simplify.

Answer 21

I'm so confused with all of this :/

Answer 22

funny how im stalking batman... very ironic xD

Answer 23

ok here it summaries up

Answer 24

sketch of what your graph would look like