I got this "parabolic" curve from a book but cannot find the right equation for it

Question

The diagram below is taken from a book on Indian Stupa architecture. It says that the profile is a "parabolic" one. I have tried y=x^2 and varied the domains ox x and y but couldn't find the right proportions. I actually don't think it is a parabolic curve. Which other function could it be? Is the grid behind of any help?

I have tried a logarithmic version of the grid but it doesn't quite fit:

This one is with the following forumla suggested by Hypergeometric:

$$\Large x=r\left(1-2^{\frac y2}\right)$$

Here is an attempt with the following equation suggested by Hypergeometric combined with an evolutionary solver to find the closest variables and it is getting much closer (domain used r(1,5.598), m(1,0.980), y(1,5.608)) :

$$\Large x=r\left(1-m^{\frac yn}\right)$$

Answer 1

It was parabolic but I should have put the exponent as a variable on y apply a scaling on the x axis i then used an evolutionary solver to find both scaling factor and exponent:

$$\Large x = t a$$

With scaling factor a = 0.17369

$$\Large y = t^b$$

With exponent b = 0.2787

The domain of t is 0 to 1

Answer 2

Interesting curve.

Try $$\Large x=r\left(1-2^{\frac y2}\right)$$ which is equivalent to $$\Large2^y=\left(1-\frac xr\right)^2$$ where $r$ is the radius or width of the asymptote. Not exactly a parabola per the classical definition, which has to fit a quadratic function.

If you want to tweak the curvature, you might want to try a generalized version of the above, as in

$$\Large x=r\left(1-m^{\frac yn}\right)$$

where $m>1$ and $n>0$. The curve approaches the asymptote more quickly as $m$ increases or as $n$ decreases. The equation given above corresponds to $m=n=2$. A good fit to the graph provided might be $m=2, n=5$.

https://www.desmos.com/calculator/isom6ksn32

NB: The above is for the right side of the curve. For the left side, replace $x$ with $-x$. The same formula won't work as the curve is not symmetrical.