Question

How to write the equation of the parabola, in vertex form, with focus at (2,-4) and directrix y = -6?

Answer 1

Do uk the vertex form equation?

Answer 2

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

Answer 3

yes

Answer 4

uh, what next?
i thought the vertex was (h,k), not the focus.

Answer 5

yea one sec

Answer 6

give me like 5mins brb

Answer 7

laughing out loud

Answer 8

HE DITCHED MEEEE

Answer 9

remember the relationship of the conics form and vertex form I showed you?

Answer 10

4p = 1/a ?

Answer 11

the whole equations

Answer 12

where is the focus and where is the vertex

Answer 13

4p(y – k) = (x – h)^2

Answer 14

p = 1/4 is the distance of the directrix on the axis of symmetry

Answer 15

directrix to the vertex

Answer 16

explain it to me like im an 8th grader pls, im dumb

Answer 17

laughing my arse off you're not

Answer 18

p = 1/4 is the distance of directrix on the axis of symmetry so..
what do i do with all this crap

Answer 19

I missed to write distance of vertex to directrix

Answer 20

let me ask you first, can you identify where does the parabola opens?

Answer 21

axis of symmetry , y = -4 ?

Answer 22

\[\huge \sqrt{(x_0 - 2)^2 + (y_0 +4)^2} = (y_0 - 2)\]
\[\huge (x_0 - 2)^2 + (y_0 +4)^2 = (y_0 - 2)^2\]
\[\huge x^2-4x+4 + y^2+8y+16 = y^2 +12y + 36\]
\[\huge y = \frac{ x^2 }{ 4 } -x - 4\]

Answer 23

ugh that should (y_0+6**

Answer 24

y + 4 = x^2/4 - x
(y+4) = x^2/4 - x
but it isn't one of the answer choices -.-

Answer 25

i could simplify it a bit more..

Answer 26

y+4 = 1/4(x-2)^2 - 1
so
4(y+5) = (x-2)^2
which is actually an answer choice, so yeah i got it right ;D

Answer 27

yo nin u dere