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December 22, 2014

December 22, 2014

Posted by **Lucy** on Sunday, March 9, 2008 at 2:07pm.

I know that you cannot graph this but the graph of this equation should be a circle I think.

I have points (3,0), (0,4), (-3, 0), and (0,-4) graphed. Is there any more points that I could use to graph?

Thanks.

- Algebra II -
**Damon**, Sunday, March 9, 2008 at 2:29pmnope, ellipse

4^2 x^2 + 3^2 y ^2 = 12^2

(4^2/12^2) x^2 + (3^2/12^2)y^2 = 1

x^2 /3^2 + y^2/ 4^2 = 1

form is

(x-h)^2/a^2 + (y-k)^2/b^2 = 1 for ellipse centered at (h,k)

so

ellipse with center at (0,0)

horizontal half length = a = 3

vertical half height = b = 4

You can put in all the intermediate points you wish, like x = 1 and x = -1 (same y answers)

and x = 2 and x = -2 (again same y answers).

- Algebra II -
**drwls**, Sunday, March 9, 2008 at 2:30pmThis equation graphs as an ellipse, not a circle. An ellipse is a kind of flattended circle.

The equation can be rewritten, after dividing both sides by 144 and recognizing some perfect squares,

(x/3)^2 + (y/4)^2 = 1

The lengths of the major and minor axes are 3x2 = 6 and 4x2 = 8 ,(twice the numbers in the denominator). The center of the ellipse is the origin, (0,0).

I suggest you plot a few more points to more accurately graph the ellipse. For example, if x = + or -1,

y = +/- 3.77

- Algebra II -
**oneisha**, Monday, November 7, 2011 at 7:25pmmajor axis 20 units long and parallel to y-axis minor axis 6 units long center at (4,2)

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