nope, 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).
Graph 16x^2 + 9y^2 = 144
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.
4 answers
This 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
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
major axis 20 units long and parallel to y-axis minor axis 6 units long center at (4,2)
y=5/x