- anonymous

can anyone explain how to do parabola's in great detail

- schrodinger

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- anonymous

yeah..fire away

- anonymous

wat d u want to know exactly?

- anonymous

study all the four standard forms and the general forms of parabolae carefully...learn how to evaluate the vertex, latusrectum and directrix and focus and centre

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## More answers

- anonymous

how to figure out the vertex directrix and focus if they only give me a lil information

- anonymous

giv me d info

- anonymous

vertex (-2,0) directrix y=-4 and i have 2 find the equation and graph it

- anonymous

(x+2)^2= 16y

- anonymous

ok now can u teach me how to get that

- anonymous

sure

- anonymous

u c if u have d directrix parallel to x-axis, its a vertical parabola
upward or dwnward

- anonymous

since here d vertex lies above d directrix its an upward parabola

- anonymous

ok so do i use (x-h)^2=4p(y-k)

- anonymous

yeah bingo...where p is the distance btween d vertex and directrix

- anonymous

got it?

- anonymous

no i just know the formula 2 use but idk how 2 use it

- anonymous

well now u kno

- anonymous

no i dont how did u get the answer wat are the steps wat did u plug in

- anonymous

u kno d formula...jst remmbr u use dis only when the directrix is parallel to x-axis

- anonymous

when d directrix is y=something, u use it

- anonymous

ok but wat steps did u take to get the answer i knoow 2 use the fromula but idk how 2 get the answer

- anonymous

see h and k are the x and y co-ordinates of the vertex. and p is the distance of the vertex from directrix

- anonymous

so far all i have is (x-2)^2=4py

- anonymous

so p=3???

- anonymous

p is 4
the distance of -2,0 from y=-4

- anonymous

so p=-4

- anonymous

4
distance is always positive

- anonymous

ok

- anonymous

so uve got (x+2)^2 = 4.4(y)

- anonymous

thanks do you mind helping me wit some more

- anonymous

no i dont..

- anonymous

(x+2)^2=16y

- anonymous

wat i got

- anonymous

thats ryt i spose

- anonymous

focus (0,0) directrix x=4

- anonymous

hold on on dat last one was my parabola suppose 2be upward and my directrix was suppose 2 go through it right???

- anonymous

no ur getting muddled up..the directrix doesnt go thru d parabola

- anonymous

well when i graphed it it did cuz the y axis is vertical and not horizontial so is my parabola not facing the right way

- anonymous

no ur parabola shld touch the x-axis, pass thru -2,0 and the directrix should be 4 units below it

- anonymous

ok i had it right the 1st time so on to the next one lol

- anonymous

ws i wrong?

- anonymous

idk lol idc 4real im sick of dis class

- anonymous

ready 2 help me wit the other 1

- anonymous

yeah

- anonymous

focus (0,0) directrix x=4

- anonymous

wat way will the directrix be going

- anonymous

vertical

- anonymous

ok

- anonymous

now 2 figure out the equation do i use (y-k)^2=4p(x-h)

- anonymous

and is this going 2 be opening to the left

- anonymous

ur right...u do it quite well..yr u sick of it?

- anonymous

y^2=-16x

- anonymous

cuz idk how 2 do it im just guessing

- anonymous

no bt ur guessin right so u shld get confidence

- anonymous

so wat is the vertex and if i dont have (h,k) do i use x,y

- anonymous

use 0,0 for h,k..so its y^2=-16x

- anonymous

ok so anytime i dont have the vertex use 0,0

- anonymous

bt ur vertex is given 0,0

- anonymous

dat was the focus

- anonymous

OH IMSORRY

- anonymous

rly sorry..bt u cn frgv a lovesick man cant u...

- anonymous

lol sure so how do i do it

- anonymous

neway so the vertex is the midpt of the focus and the pt where d directrix meets the axis

- anonymous

i need numbers to do the formula

- anonymous

so its d midpt of 0,0 and 4,0

- anonymous

so the vertex is 2,0

- anonymous

oooooo ok but y lol

- anonymous

ur having fun arent u...d midpt of 0,0 and 4,0 is (0+4)/2, 0+0/2

- anonymous

vertex is d midpt of the focus and the directrix meeting the axis

- anonymous

u should be my teacher u making it simple my teacher is making it hard ok so i can always use dat when i am givin the focus and directrix right

- anonymous

is the answer y^2=-8px

- anonymous

thnx fr d compliment..wt grade r u in?

- anonymous

11th

- anonymous

im in ap math

- anonymous

nahi...replace d x with (x-2)

- anonymous

cz d vertex is 2,0

- anonymous

so y^2=4p(x-2)

- anonymous

yeah

- anonymous

ok dis one i did they gave me the center (-7,3) and the focus (-7,5) i got (x+7)^2=-24y is dat right

- anonymous

(x+7)^2 = 8(y-3)

- anonymous

ok

- anonymous

now the other problems r harder they gave me just a graph wit the drawing already made so i hope i can explain dis 2 u

- anonymous

try

- anonymous

the is (-4,3) and the directrix is (7,2.5)

- anonymous

ur doing somthin wrng

- anonymous

the vertex was the (-4,3)

- anonymous

is it (x-3)^2=4p(y+4)

- anonymous

yeah

- anonymous

ok

- anonymous

ok the vertex is (-2,_6) and they gave me 2.25 wat do i do

- anonymous

(-2,-6)

- anonymous

whts dis 2.25

- anonymous

x=2.25

- anonymous

so its (y+6)^2 = - 17(x+2)

- anonymous

where u get -17

- anonymous

4.25 x 4

- anonymous

where u get 4.25

- anonymous

distance of -2 from 2.25

- anonymous

u lost me now i have (y=6)^2=4p(2.25+2)

- anonymous

(y+6)^2

- anonymous

= 4.25(4) (x+2)

- anonymous

so p=4.25

- anonymous

yeah right

- anonymous

but how u know to do 2.25+2 then to make dat p

- anonymous

remmbr its d distance between d vertex and directrix

- anonymous

now im all lost x=2.25 is the directrix right

- anonymous

yes
and x=-2 is the vertex so d distance of it from 2.25 is 4.25

- anonymous

my bad the directrix is x=-2.5

- anonymous

x=-2.25

- anonymous

so its only .25

- anonymous

its from -2 to 2.25 so it 2.25 - (-2) = 4.25

- anonymous

no if vertex is -2 and the directrix is -2.25 there only 1/4 apart

- anonymous

am i wrong

- anonymous

if directrix is 2.25 then yes

- anonymous

no the directrix is negative

- anonymous

so wats the new answer

- anonymous

then ur right

- anonymous

so (y+6)^2=p(x+2)

- anonymous

yeah

- anonymous

yay thank u so much

- anonymous

r u good at doing dis wit word problems

- anonymous

nvm if i get stuck ill post it but thanks 4 all ur help

- anonymous

no prob

- anonymous

can u help me

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