Why does the graph of $y^2=1-\frac{4x^{10^{12}}}{\pi^2}$ look so much like a square? 
I want to know why the equation $y^2=1-\dfrac{4x^{10^{12}}}{\pi^2}$ gives an approximate square. (See the figure below.)


Background
I was just playing around with functions and I wanted to see if $y=\left|\sin\bigg(\dfrac{\pi x}{2}\bigg)\right|$ (radians) would give a semicircle for the interval $[0,2]$ as the distance of $(1,0)$ is the same from $(0,0)$, $(2,0)$ and $(1,1)$, all of which will lie on the curve. The equation of a unit semicircle with its centre at $(1,0)$ is $y=\sqrt{2x-x^2}$.
I know that the curves of both the equations don't resemble each other much but I still thought of approximating the sine function using this because I thought that it could still be combined with another approximation to make a better approximation. Anyway, I did it and for $\phi=x~\mathrm{radians}$, the value of $\sin\phi$ can to be approximately $\dfrac2\pi\sqrt{\pi x-x^2}$. It looked like a semi-ellipse and so I verified it to find that it was a semi-ellipse. I thought of using this to derive the equation for an ellipse with it's centre at the origin and the value of $a$ and $b$ being $\dfrac\pi2$ and $1$ respectively.
The equation came out to be : $y^2 = 1 - \dfrac{4x^2}{\pi^2}$

Finally, I thought of playing with this equation and changed the exponent of $x$. I observed that as I increased the power, keeping it even, the figure got closer and closer to a square.
$y^2=1-\dfrac{4x^{10^{12}}}{\pi^2}$ gave a good approximation of a square. For the exponent of $x$ being some power of $10$ greater than $10^{12}$, a part of the curve began to disappear.

I want to know why this equation gives an approximate square.
Note : I would like to inform you that I have no experience with conic sections.
Thanks!
 A: This is a rectangle, because for $x=0$ we get $|y|=1$, but for $y=0$ we obtain
$$
x=\root{10^{12}}\of{\pi^2\over4}\approx
1.0000000000009031654.
$$
For a square, you'd better replace ${4\over\pi^2}$ with $1$.
A: HINT
$y=\pm 1$ is clearly a tendency around $x=0$ and the
$y=\log[(4/\pi)^2  x^{m}] $  tends to pass through $(x=1, x=-1)$ as  $y\rightarrow 0$
A: This is related to what happens with the graphs of very high powers of $x,$
which in turn is related to exponential growth and decay.
Graph $y = x^2.$ Notice that the curve goes through $(0,0)$ at its low point, and goes through $(-1,1)$ on the left and $(1,1)$ on the right.
And the graph has a tiny nearly level section very near the bottom.
Try $y = x^4.$ It's somewhat like $y=x^2$, but the sides are steeper at
$(-1,1)$ and $(1,1)$ and the bottom is much flatter.
Try $y = x^{10}$. Steeper sides, flatter bottom than $x^4.$
As you try higher and higher powers of $x,$ you get a larger and larger "flat" part at the bottom of the curve.
This part isn't really flat, it's just that for any number $x$ with $|x|<1,$
if you look at $x^n$ and increase the exponent $n$ you have a process of exponential decay where $x^n$ approaches zero. At some exponent the value of $x^n$ will be so small that you cannot see the difference between $x^n$ and zero on the graph.
For values of $x$ closer to $\pm 1$, $x^n$ decays slower and it takes a higher value of $n$ before $x^n$ gets close enough to zero to be indistinguishable from zero by your eye. But if you take really large values of $n$, such as $10^{12},$ the numbers near $\pm1$ for which $x^n$ is not visually indistinguishable from zero are so close to $\pm1$ that they are visually indistinguishable from $1$ and the graph looks like it has straight vertical sides there.
In fact even at $n = 1000$ the graph looks pretty square at the bottom to me.
Now flip the graph over by taking $y = 1 - x^n$ for a very large value of $n.$
It still has that rectangular shape, but the flat level part is at $y = 1$
and the rest is below that, passing through $(-1,0)$ and $(1,0)$.
Now take $y = \sqrt{1 - x^n}.$ If $n$ is large enough this still looks rectangular, but the parts of the graph below the $x$ axis have disappeared because negative numbers do not have real square roots.
If you now square both sides, $y^2 = {1 - x^n},$
you get the same result above the $x$ axis,
but since $(-y)^2 = y^2$ you get two symmetric values of $y$ for each value of $x,$
that is, the graph above the $x$ axis is mirrored below the $x$ axis,
forming what looks like a square.
Multiplying $x^n$ by some positive constant $a$, as in $y^2 = {1 - ax^n},$
makes the graph wider or narrower in the $x$ direction.
That is, you are graphing $y^2 = {1 - (a^{1/n}x)^n},$
so the graph is scaled by a factor of $a^{-1/n}$ in width.
If $a$ is not too large (for example, $a = 4/\pi^2$) and $n$ is very large,
$a^{-1/n}$ is extremely near $1$ (as other answers have pointed out).

For the exponent of $x$ being some power of $10$ greater than $10^{12}$, a part of the curve began to disappear.

I had a similar experience with extremely high powers of $x$, using the graphing calculator at Desmos.com. I suspect this is a limitation of the size of number that the calculator can deal with, or perhaps the horizontal step size (graph so steep that the software cannot increment $x$ slowly enough to plot a continuous curve).
