What are the foci of the ellipse #x^2/49+y^2/64=1#?

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Answer 1 · 2015-03-02T11:21:17

The answer is: #F_(1,2)(0,+-sqrt15)#.

The standard equation of an ellipse is:

#x^2/a^2+y^2/b^2=1#.

This ellipse is with the foci (#F_(1,2)#) on the y-axis since #a < b#.

So the #x_(F_(1,2))=0#

The ordinates are:

#c=+-sqrt(b^2-a^2)=+-sqrt(64-49)=+-sqrt15#.

So:

#F_(1,2)(0,+-sqrt15)#.