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I need to graph the following surface: $4x^2 + y^2 = 4$ which is an ellipse, so canonical equation is $x^2 + (y^2)/4 = 1$. Then, to graph this I can say:

If $x=0$:

$y^2 = 4$

so $y = 2$ or $y = -2$

But what I have in the plane? two parallels lines? because I just see two points at that coords in the following graph:

click me

JMP
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Roberto Sepúlveda
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    Could you possibly expand a bit on your question? At $x=0$ you have correctly identified that $y=\frac+{}2$, what's the problem? – Aka_aka_aka_ak Apr 03 '17 at 19:37
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    @Aka_aka_aka_ak, you can use $\pm$ for $\pm$. –  Apr 03 '17 at 19:41
  • @tilper oh thanks – Aka_aka_aka_ak Apr 03 '17 at 19:44
  • What is the question, exactly? – mlc Apr 03 '17 at 19:46
  • @Aka_aka_aka_ak Can I say there's two paralel lines if I don't see that lines on the plane? – Roberto Sepúlveda Apr 03 '17 at 19:48
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    Well, that graph is only one crosssection of the entire surface that in 3D space is a tube. If you expand those two points into the front and back third dimension they will expand into two parallel lines just like the ellipse will expand into a tube. – fleablood Apr 03 '17 at 19:48
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    The lines aren't on the xy plane. They are on the yz plane that goes from your eyeballs to the screen. The two lines are poking two holes into your computer screen and piercing you skull and going out the back of your house. – fleablood Apr 03 '17 at 19:50

2 Answers2

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If you are graphing in xyz space the surface is a tube with a constant x/y cross section being an ellipse for any $z = k$.

If you take the y/z cross section at $x = 0$ this is cutting the tube lengthwise in half and seeing the cut lines. The result is two parallel lines.

That's exactly what you should expect.

If you are graphing only on the x-y plane the this is an ellipse. Setting $x = 0$ gives you the two points on the ellipse where $x =0$. That there are two points and they are $(0,2)$ and $(0,-2)$ should not surprise any one.

fleablood
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You get parallel lines when $y= -2$ or $ y = $ any constant $c$ FOR EVERY $x$. Please be careful, in the context or your question its only the case that $y= 2$ or $y = -2$ when $x=0$ not at any other points.

Ziad Fakhoury
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