SOLUTION: : how would you find the foci of this ellipse? x^2+9y^2=1

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Question 194144This question is from textbook
: : how would you find the foci of this ellipse?
x^2+9y^2=1 This question is from textbook

Found 2 solutions by nerdybill, stanbon:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
.
Equation for any ellipse:
+%28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1
.
So, we can rewrite:
x%5E2%2B9y%5E2=1
As:
x%5E2+%2B+y%5E2%2F%281%2F9%29+=+1
x%5E2%2F1%5E2+%2B+y%5E2%2F%281%2F3%29%5E2+=+1
%28x-0%29%5E2%2F1%5E2+%2B+%28y-0%29%5E2%2F%281%2F3%29%5E2+=+1
.
From the above we now know:
(h,k) = (0,0)
a = 1
b = 1/3
.
a^2 - b^2 = c^2
1^2 - (1/3)^2 = c^2
1 - 1/9 = c^2
8/9 = c^2
sqrt(8/9) = c
(2/3)sqrt(2) = c
.
And, since 'a' is larger than 'b' -- it is horizontal major
therefore, foci is at
(h+-c, k)
or
(0+-(2/3)sqrt(2), 0 )
.
Foci would then be:
((2/3)sqrt(2), 0 )
and
(-(2/3)sqrt(2), 0 )

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
find the foci of this ellipse?
x^2+9y^2=1
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Put the equation in standard form:
[x^2/1] + [y^2/(1/9)] = 1
The center is at (0,0)
The foci are at (-c,0), (c.0)
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Then a = 1 and b = 1/3
----------
For an ellipse a^2 = b^2 + c^2
So c^2 = 1^2 - (1/3)^2 = 8/9
Then c = (2/3)sqrt(2)
---------------------------
Foci are at ((-2/3)sqrt(2)),0) and ((2/3)sqrt(2),0)
=============================================
Cheers,
Stan H.