Is the graph of (x^2/4) + y^4 = 1 an ellipse?

Question

My first thought was that it was not an ellipse because it was not because y was not to the 2nd power. But I decided to graph it to find out. I graphed the following four equations: y = +/-[(+/-) (1-(x^2/4))^1/2]^1/2. I got 2 imaginary values for y and 2 real values for y. The graph of the function was an ellipse with a=2 and b=1. Is this the correct solution? Thank you.

T. Grode

Your first thought was right. The graph does not have all the properties of ellipse. — z100, May 21 '17 at 17:29
Maybe of interest https://en.wikipedia.org/wiki/Superellipse and http://mathworld.wolfram.com/Superellipse.html — cgiovanardi, May 22 '17 at 03:32

Fly by Night · Accepted Answer · 2017-05-22T17:54:27.927

It is not an ellipse.

If it were a circle (a special case of an ellipse) then you'd be able to find a point in the plane $P$ (the centre) such that the distance from $P$ to every point $X$ on the curve, written $|XP|$, is constant.

If it were a "proper" ellipse then you would be able to find two, and only two, points in the plane $P_1$ and $P_2$ (the foci) such that $|XP_1|+|XP_2|$ is constant for all $X$.

http://www.mathwords.com/e/ellipse.htm

score 0 · Answer 2 · answered May 21 '17 at 17:31

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I used WolframAlpha to graph $$\frac{x^2}{4}+y^4=1$$ and the resulting plot is shown in the image below:

To cut the long story short, this is certainly not an ellipse.

answered May 21 '17 at 17:31

Jose Arnaldo Bebita Dris

1

Is the graph of (x^2/4) + y^4 = 1 an ellipse?

2 Answers2