help me plz

- anonymous

help me plz

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

##### 1 Attachment

- anonymous

@Directrix

- anonymous

brb in 20

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

im back

- Nnesha

first of all you should know is it horizontal or vertical
a always greater than b `a>b`
if a is under the x variable then it would be `horizontal equation`
\[\rm \frac{ (x-h)^2 }{\color{ReD}{ a^2} }+ \frac{ (y-k)^2 }{ b^2 }=1\]
if a is under the y variable then it would be `vertical equation`
\[\rm \frac{ (x-h)^2 }{ b^2 }+ \frac{ (y-k)^2 }{\color{reD}{ a^2} }=1\]

- Nnesha

so is it vertical or horizontal ?

- anonymous

i think its vertical

- Nnesha

foci for horizontal \[\large\rm (h, k \pm c)\]
and for vertical equation \[\large\rm (h \pm c , k)\]
use the equation to find c \[\huge\rm c^2 =a^2-b^2\]

- anonymous

ok so which one do i use

- Nnesha

well a is bigger number which is under the x variable so is it vertical or horizontal ?

- Nnesha

read my first comment :=)

- anonymous

so it is horizontal right

- Nnesha

yes right \[\huge\rm \frac{ x^2 }{ \color{ReD}{36} } +\frac{ y^2 }{ 11 }=1\] now use the equation posted above to find c

- Nnesha

c^2=a^2-b^2 this one

- anonymous

wait what

- Nnesha

foci for horizontal \[\large\rm (h, k \pm c)\]
and for vertical equation \[\large\rm (h \pm c , k)\]
those two are foci pair
we need
so
use this equation to find c \[\huge\rm c^2 =a^2-b^2\]

- Nnesha

a^2 and b^2 are in the given equation

- anonymous

wow ok so umm can you start me off plz

- Nnesha

`horizontal equation` for ellipse
\[\rm \frac{ (x-h)^2 }{\color{ReD}{ a^2} }+ \frac{ (y-k)^2 }{ b^2 }=1\]
the number at the denominator are a^2 and b^2
and like i said bigger number equal to a^2

- Nnesha

what two numbers are replaced with a^2 and b^2 in this equation \[\huge\rm \frac{ x^2 }{ \color{ReD}{36} } +\frac{ y^2 }{ 11 }=1\] ??

- anonymous

brb i gtg cook dinner

- anonymous

36 and 11

- Nnesha

right which one is a^2 and which one is b^2 ?

- anonymous

a^2 is 36 b^2 is 11

- anonymous

@IrishBoy123

- anonymous

{ (x-h)^2 }{36} }+/{ (y-k)^2 }{11}=1 is this the correct formula

- IrishBoy123

this ellipse is centred on the origin
yep?

- anonymous

yep

- IrishBoy123

|dw:1444085734817:dw|

- anonymous

yep

- IrishBoy123

Para-phrasing Wiki:
The distance from the origin O to either focus is f where:
\[f = \sqrt{a^2-b^2}.\]

- anonymous

ok she gave me a diffrent formula soo what do i do im so confused now

- IrishBoy123

here, what is a and what is b?
general equation for ellipse centred on Origin is:
\(\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1\)

- IrishBoy123

a = ??
b = ??

- IrishBoy123

then
|dw:1444086259207:dw|

- anonymous

(x-h)^2 }{36} }+/{ (y-k)^2 }{11}= ok so i got this

- IrishBoy123

so that means you have finished it?!

- anonymous

no i have not thats the first equation

- anonymous

(y-k)^2 {11} this is my answer

- IrishBoy123

OK
for an equation centred on O, we have general equation:
\[\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1\]
The focii are at a distance f from O, where
\[f = \sqrt{a^2-b^2}\]
|dw:1444086808815:dw|

- IrishBoy123

soz, we crossed posts.
i understood the question to be finding the focii of the ellipse

- anonymous

yes idk what im doing im so cunfused here im sorrry

- IrishBoy123

no worries
you have here: \(\large\rm \frac{ x^2 }{ \color{ReD}{6^2} } +\frac{ y^2 }{(\sqrt{\color{red} 1\color{red}1 })^2}=1\)
and generally speaking: \(\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1\)
and: \(\large f = \sqrt{a^2-b^2}\)

- IrishBoy123

so f =????

- anonymous

omg i have no clue on how to do this like wth

- anonymous

does f equal -4

- IrishBoy123

\(a = 6, \; b = \sqrt{11}\)
\(f = \sqrt{ 6^2 - (\sqrt{11})^2}\)
\(f = 5\)

- anonymous

atleast i was close haha

- IrishBoy123

finished ??
:p

- anonymous

i think so

- IrishBoy123

cool!! well done

- anonymous

thnx for your help

- IrishBoy123

mp

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @Tazmaniadevil
ok she gave me a diffrent formula soo what do i do im so confused now
\(\color{blue}{\text{End of Quote}}\)
the formula is same
difference is f and c variables
f= foci
\[\huge\rm f = \sqrt{ a^2 -b^2}\]
is same as \[\huge\rm c^2= a^2-b^2 ~~~~~c=\sqrt{a^2-b^2}\]

- Nnesha

a^2 and b^2 are already given
so would be easy you write it as c= sqrt{36-11} :=)

- anonymous

OK

Looking for something else?

Not the answer you are looking for? Search for more explanations.