anonymous
  • anonymous
help me
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@LynFran
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Nnesha
  • Nnesha
is it horizontal or vertical ?
anonymous
  • anonymous
??

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Nnesha
  • Nnesha
is it horizontal ellipse or vertical ellipse?
anonymous
  • anonymous
idk lol
LynFran
  • LynFran
c^2=a^2-b^2
Nnesha
  • Nnesha
if bigger number under the x coordinate then horizontal and if it's under the y coordiante then vertical
anonymous
  • anonymous
so its horizontal
Nnesha
  • Nnesha
yes right so thats mean you should add c into x coordinate of the center \[\huge\rm (h \pm c,k)\] now use that equation lynn gave u to find c value
Nnesha
  • Nnesha
(h,k) is the center
anonymous
  • anonymous
ok
Nnesha
  • Nnesha
btw standard form of ellipse is \[\huge\rm \frac{ (x-h)^2 }{ a^2 } + \frac{ (y-k)^2 }{ b^2 }=1\] a^2= ?? b^2= ?? in ur question
anonymous
  • anonymous
yeah how do i found this out
Nnesha
  • Nnesha
they already in the question a^2 is under the x b^2=under the y
anonymous
  • anonymous
ok sooooo
Nnesha
  • Nnesha
substitute a^2 and b^2 value into the equation lynn gave you c^2=a^2-b^2 then solve for c
anonymous
  • anonymous
so hard im just now learing this lol
anonymous
  • anonymous
ohh
Nnesha
  • Nnesha
thats fine \[\huge\rm \frac{ (x-h)^2 }{ \color{ReD}{a^2 }} + \frac{ (y-k)^2 }{\color{ReD}{ b^2 }}=1\] \[\huge\rm \frac{ x^2 }{\color{Red}{ 36} }+\frac{ y^2 }{\color{ReD}{ 11} }=1\] bg number would be `a^2` always!(in ellipse equation)
Nnesha
  • Nnesha
so a^2 = ?
anonymous
  • anonymous
1
Nnesha
  • Nnesha
no a^2 is under the x^2 coordinate
Nnesha
  • Nnesha
http://prntscr.com/89k6q9
anonymous
  • anonymous
so36
Nnesha
  • Nnesha
yes right now what is b^2=?
anonymous
  • anonymous
11
Nnesha
  • Nnesha
perfect! \[\huge\rm c^2=a^2-b^2\] substitute a^2 and b^2 for their values solve for c
anonymous
  • anonymous
c^2=-1175
anonymous
  • anonymous
am i right
Nnesha
  • Nnesha
ayooo how :o
Nnesha
  • Nnesha
a^2=36 b^2=11 it's already squared you don't need to square them again
anonymous
  • anonymous
sooo
Nnesha
  • Nnesha
so.. substitute a^2 for 36 and b^2 for 11
anonymous
  • anonymous
25
anonymous
  • anonymous
so c^2=25
Nnesha
  • Nnesha
yes right now take square root both sides to cancel out the square
anonymous
  • anonymous
so c=5
Nnesha
  • Nnesha
yes right now what's the center point ?
anonymous
  • anonymous
idk
Nnesha
  • Nnesha
\[\huge\rm \frac{ (x-\color{reD}{h})^2 }{ a^2 } + \frac{ (y-\color{ReD}{k})^2 }{ b^2 }=1\] are there any h and k value at the numerator ?
anonymous
  • anonymous
??
Nnesha
  • Nnesha
look at the original equation is there any number with x ?
anonymous
  • anonymous
yeah its sqared
Nnesha
  • Nnesha
not square (x-3)^2 then h would be 3 for this equation center point is (0,0)
Nnesha
  • Nnesha
so formula for foci is \[\huge\rm (h \pm c,k)\] where (h,k) is the center (0,0) c=5 substitute variables for values

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