Posted by Soly on Thursday, November 22, 2007 at 7:48pm.
I am not too sure about this problem. Can you please help me?
Graph the ellipse and locate the foci.
9x^2=14416y^2
I got ( squareroot 7, 0) and (squareroot 7,0), but I was looking at one problem in one book similar to it, and it had the answer as (0,  squareroot 7) and (0, squareroot 7).

Algebra  Reiny, Thursday, November 22, 2007 at 10:15pm
You are correct, I got the same foci
The one you looked at in your text probably had the yaxis as its major axis, that is, the focal points were on the y axis.
Was the denominator of the x^2 smaller than the denominator of the y^2 term?
If so, that was the case.

Algebra  Soly, Thursday, November 22, 2007 at 10:20pm
The one in book, is 7x^2=355y^2

Algebra  drwls, Thursday, November 22, 2007 at 10:34pm
Rewrite as
x^2/(144/9) + y^2/(144/16) = 1
(x/4)^2 + (y/3)^2 = 1
The semimajor and semiminor axis distances are 4 and 3, respectively. The major axis lies along the x axis. The distance from the center to the foci is sqrt (4^2  3^2) = sqrt 7
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