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Mathematics 21 Online
OpenStudy (anonymous):

locate the foci of the ellipse

OpenStudy (anonymous):

Nnesha (nnesha):

first of all ) is it horizontal or vertical ?

Nnesha (nnesha):

a > b if a is under the x variable then it's horizontal and if it's under the y variable then it would be vertical :=))

OpenStudy (anonymous):

horizontal

Nnesha (nnesha):

right foci for horizontal ellipse \[\huge\rm (h \pm c ,k)\] where (h,k) is the center to find value of use the equation \[\huge\rm c^2 = a^2-b^2\] plugin a and b value solve for c and then add/subtract c value into x-coordinate of the center

OpenStudy (anonymous):

can u help me nnesha when ready i just need checking

Nnesha (nnesha):

alright let me know if you have any question osama :=))

OpenStudy (anonymous):

\[c=\sqrt{1175}\]

OpenStudy (anonymous):

c = 34.3

Nnesha (nnesha):

hmm did you take square of a 36^2 ???

Nnesha (nnesha):

its already squared at the denominator \[\frac{ (x-h)^2}{ a^2 }+\frac{{(y-k)^2}}{b^2}=1\] standard form equation

Nnesha (nnesha):

a^2 = 36 b2 =11

Nnesha (nnesha):

b^2 **

OpenStudy (anonymous):

i'm confused now, i thought i had to plug in c^2=a^2-b^2 c^2=36^2-11^2 c^2=1296-121

OpenStudy (anonymous):

i take it that's wrong then

Nnesha (nnesha):

look at the general form equation of a circle a and b already squared a =6 a^2 = 36 |dw:1445027278050:dw|

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