How do you get hyperbola by squaring a bell? This is a Gaussian bell (aka normal distribution).

Its square, I belive looks the same. Yet, I see that chi-square distribution, which is a sum of k such bell squares, looks like 

Take a look at yellow chi-squared, with k=1. It should be just single bell squared. But, it looks like a hyperbola instead of Gaussian. Why?
PS I read that normally-distributed signal generates a uniformly distributed power. Is it the same kind of magic? 
 A: Consider a random variable $X$ with density $f$, say $X$ gaussian, and $Y=X^2$. Then $Y\geqslant0$ almost surely and, for every $y\geqslant0$, $[Y\leqslant y]=[-\sqrt{y}\leqslant X\leqslant\sqrt{y}]$ hence
$$
P[Y\leqslant y]=\int_{-\sqrt{y}}^{\sqrt{y}}f(x)\mathrm dx.
$$
Differentiating this yields the density of $Y$, namely, for every $y\geqslant0$,
$$
g(y)=\frac{f(\sqrt{y})+f(-\sqrt{y})}{2\sqrt{y}}.
$$
When $X$ is normal (centered or not), $f$ is continuous at $0$ hence the behaviour of $g$ at $0$ is
$$
g(y)\sim\frac{f(0)}{\sqrt{y}}.
$$
Note finally that $f^2$ is rarely a density function.
Edit: Recall that an integral depending on a parameter, say
$$
J(y)=\int_{u(y)}^{v(y)}w(x)\mathrm dx,
$$
has derivative
$$
J'(y)=w(v(y))v'(y)-w(u(y))u'(y).
$$
A: Let $Z$ be standard normal, or more generally a normal with mean $0$. But in fact we will use almost no properties of the normal. 
Let $Y=Z^2$. We want to show that the density function $f_Y(y)$ of $Y$ is very large for $y$ near $0$. This density function is approximately $\frac{F_Y(y)}{y}$. 
We have $F_Y(y)=2\Pr(Z\le \sqrt{y})$. For $y$ close to $0$, the density function of $Z$ is close to a non-zero constant $c$. 
Thus $F_Y(y)$ is approximately $2c\sqrt{y}$. It follows that near $0$, $\frac{F_Y(y)}{y}$ is very large. 
