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