anonymous
  • anonymous
help me plz
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
@Directrix
anonymous
  • anonymous
brb in 20

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anonymous
  • anonymous
im back
Nnesha
  • 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
  • Nnesha
so is it vertical or horizontal ?
anonymous
  • anonymous
i think its vertical
Nnesha
  • 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
  • anonymous
ok so which one do i use
Nnesha
  • Nnesha
well a is bigger number which is under the x variable so is it vertical or horizontal ?
Nnesha
  • Nnesha
read my first comment :=)
anonymous
  • anonymous
so it is horizontal right
Nnesha
  • 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
  • Nnesha
c^2=a^2-b^2 this one
anonymous
  • anonymous
wait what
Nnesha
  • 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
  • Nnesha
a^2 and b^2 are in the given equation
anonymous
  • anonymous
wow ok so umm can you start me off plz
Nnesha
  • 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
  • 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
  • anonymous
brb i gtg cook dinner
anonymous
  • anonymous
36 and 11
Nnesha
  • Nnesha
right which one is a^2 and which one is b^2 ?
anonymous
  • anonymous
a^2 is 36 b^2 is 11
anonymous
  • anonymous
@IrishBoy123
anonymous
  • anonymous
{ (x-h)^2 }{36} }+/{ (y-k)^2 }{11}=1 is this the correct formula
IrishBoy123
  • IrishBoy123
this ellipse is centred on the origin yep?
anonymous
  • anonymous
yep
IrishBoy123
  • IrishBoy123
|dw:1444085734817:dw|
anonymous
  • anonymous
yep
IrishBoy123
  • IrishBoy123
Para-phrasing Wiki: The distance from the origin O to either focus is f where: \[f = \sqrt{a^2-b^2}.\]
anonymous
  • anonymous
ok she gave me a diffrent formula soo what do i do im so confused now
IrishBoy123
  • 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
  • IrishBoy123
a = ?? b = ??
IrishBoy123
  • IrishBoy123
then |dw:1444086259207:dw|
anonymous
  • anonymous
(x-h)^2 }{36} }+/{ (y-k)^2 }{11}= ok so i got this
IrishBoy123
  • IrishBoy123
so that means you have finished it?!
anonymous
  • anonymous
no i have not thats the first equation
anonymous
  • anonymous
(y-k)^2 {11} this is my answer
IrishBoy123
  • 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
  • IrishBoy123
soz, we crossed posts. i understood the question to be finding the focii of the ellipse
anonymous
  • anonymous
yes idk what im doing im so cunfused here im sorrry
IrishBoy123
  • 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
  • IrishBoy123
so f =????
anonymous
  • anonymous
omg i have no clue on how to do this like wth
anonymous
  • anonymous
does f equal -4
IrishBoy123
  • IrishBoy123
\(a = 6, \; b = \sqrt{11}\) \(f = \sqrt{ 6^2 - (\sqrt{11})^2}\) \(f = 5\)
anonymous
  • anonymous
atleast i was close haha
IrishBoy123
  • IrishBoy123
finished ?? :p
anonymous
  • anonymous
i think so
IrishBoy123
  • IrishBoy123
cool!! well done
anonymous
  • anonymous
thnx for your help
IrishBoy123
  • IrishBoy123
mp
Nnesha
  • 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
  • Nnesha
a^2 and b^2 are already given so would be easy you write it as c= sqrt{36-11} :=)
anonymous
  • anonymous
OK

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