SOLUTION: What are the foci of the ellipse x^2/16+y^2/49=1?

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Question 332784: What are the foci of the ellipse x^2/16+y^2/49=1?
Answer by jrfrunner(365) About Me  (Show Source):
You can put this solution on YOUR website!
What are the foci of the ellipse x%5E2%2F16%2By%5E2%2F49=1?
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x%5E2%2F16%2By%5E2%2F49=1 can be rewritten as x%5E2%2F4%5E2%2By%5E2%2F7%5E2=1
or in general
%28x-h%29%5E2%2Fb%5E2%2B%28y-k%29%5E2%2Fa%5E2=1 for an ellipse with major axis along the x direction (note: b%5E2%3Ea%5E2)
%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1 for an ellipse with major axis along the y direction (note: b%5E2%3Ea%5E2)
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x%5E2%2F4%5E2%2By%5E2%2F7%5E2=1
is an ellipse with the major axis along the y direction b%5E2=7%5E2+%3E+a%5E2=4%5E2 and the minor axis along the x direction
centered at (h,k)=(0,0)
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In ellipses the focal length is 2*c, where c = sqrt%28b%5E2-a%5E2%29+=+sqrt%2849-16%29=5.745
so the foci for an ellipse with major axis along the y direction = (h,k-c), (h,k+c)
since h=0, k=0, c=5.745, the foci are at (0,-5.745) and (0,5.745)