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January 30, 2015

January 30, 2015

Posted by **Soly** on Thursday, November 22, 2007 at 7:48pm.

Graph the ellipse and locate the foci.

9x^2=144-16y^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 10:15pmYou are correct, I got the same foci

The one you looked at in your text probably had the y-axis 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 10:20pmThe one in book, is 7x^2=35-5y^2

- Algebra -
**Reiny**, Thursday, November 22, 2007 at 10:45pm7x^2=35-5y^2 rewritten in standard form is

x^2/5 + y^2/7 = 1

sure enough the y term denominator is 7 which is larger than the 5 under the x^2.

Which means the y axis is the major axis like I said before and the focal points are on the y-axis

- Algebra -
- Algebra -
**drwls**, Thursday, November 22, 2007 at 10:34pmRewrite 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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