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is it horizontal or vertical ?
??

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