Proving a round shape is ellipse

Question

The question background is physical but my question is mathematical, or I think so.

When a circular disk with radius r in it's resting frame $S'$ moves with velocity $V$ in observer's resting frame $S$ is under the effect of length contraction.

Now the math part.

Radius that is parallel to $V$ is $x'=\frac{1}{\gamma}x, (x'<x)$, and $y'=y$. And radius in between is $r'=\sqrt{\frac{1}{\gamma ^2}x^2 + y^2}$

Intuitively, I think this is an ellipse. But I don't know how to show it. My knowledge about ellipse is pretty bad, I am thinking calculate the foci distance and show it by definition. Is there another way to do this?

Answer 1

Since$$\sqrt{\frac1{\gamma^2}x^2+y^2}=r'\iff\frac{x^2}{\gamma^2}+y^2=(r')^2\iff\frac{x^2}{(\gamma r')^2}+\frac{y^2}{(r')^2}=1,$$you have the standard expression of an ellipsis.

Answer 2

Standard equation of a circle is

$$ \frac{x^2}{ a^2} +\frac {y^2}{ a^2} =1 $$

After contraction along $x-$ axis, ellipse major axis becomes for $\gamma>1$

$$ \frac{x^2 \gamma^2}{ a^2} +\frac {y^2}{ a^2} =1 $$

which is in its standard form.