locate the foci of the ellipse
first of all ) is it horizontal or vertical ?
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 :=))
horizontal
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
can u help me nnesha when ready i just need checking
alright let me know if you have any question osama :=))
\[c=\sqrt{1175}\]
c = 34.3
hmm did you take square of a 36^2 ???
its already squared at the denominator \[\frac{ (x-h)^2}{ a^2 }+\frac{{(y-k)^2}}{b^2}=1\] standard form equation
a^2 = 36 b2 =11
b^2 **
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
i take it that's wrong then
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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