here is my proof
Formula: [4b+a^2(π-2)/b+a(π-2)]
Let A minor axis Let B major axis
1.)Example: given (A= 0) (B= 1000)
Solving: [4*1000+0^2(π-2)/1000+0(π-2)] Answer= 4000 because A is 0 so the answer is 4000 or only value of B will remain 1000 and no more point something
2.)Example: given (A= 1000) (B= 1000)
Solving:[4*1000+1000^2(π-2)/1000+1000(π-2)] Answer= 6283.185308
because the A and B are the same so the answer is like full circle circumference
3.)Example: given (A=1) (B= 1000)
Solving:[4*1000+1^2(π-2)/1000+1(π-2)] Answer= 4001.142734247
The answer is close to 4000....
Let's first look at the circle for inspiration.
A circle with radius $r$ at the origin is defined up to translation by $\frac{x^2}{r^2} + \frac{y^2}{r^2} = 1$
so $x = \sqrt{r^2 - y^2}$ well-defines the circle in the first quadrant, and we calculate $$\dfrac{dx}{dy} = - \frac{y}{\sqrt{r^2 - y^2}}$$ We'll do an arclength integral in cartesian coordinates and then multiply by $4$ for symmetry.
The length of the arc (I hope you've seen this before) is $$\int\limits_{y=0}^{y=r} dy \sqrt{1 + \big(\frac{dx}{dy}\big)^2} = r\int\limits_{y=0}^{y=r} dy \sqrt{\frac{1}{r^2 - y^2}}$$
You can recognize the expression $\sqrt{\frac{1}{r^2 - y^2}}$ as the derivative of $\arcsin(y/r)$, and thus the integral evaluates to $$r (\arcsin(r/r) - \arcsin(0/r) = \frac{\pi r}{2} $$
As mentioned before, this was just the top right quarter of the circle, so multiplying by four gives us the actual perimeter of the circle, $2 \pi r$.
So what about the ellipse??
It's pretty much the same deal! Now our ellipse, up to translation, is defined by $(\frac{x}{a})^2 + (\frac{y}{b})^2 = 1$. By considerations of symmetry we can again focus on a single quadrant, in which $x = \frac{a}{b} \sqrt{b^2 - y^2}$, and $$\frac{dx}{dy} = -\frac{a}{b} \frac{y}{\sqrt{b^2 - y^2}}$$
Now the integral ranges from $y = 0$ to $y = b$, and you can set it up just like above, as
$$\text{Perimeter of Ellipse} = 4*\int\limits_{y=0}^{y=b} dy \sqrt{1 + \big(\frac{dx}{dy}\big)^2}$$
You'll see the integral is pretty messy!
About the best you can do here for closed form is to fiddle with a linear change of variable and define the eccentricity constant $e^2 = 1 - \frac{a^2}{b^2}$, to put this into the form
$$\text{Perimeter of Ellipse} = 4*b*\int\limits_{y=0}^{y=1} dy \sqrt{ \frac{1-e^2y^2}{1 - y^2}}$$ The integral expression here is known as a complete elliptic integral of the second kind.
Note that when the eccentricity is $0$, the minor and major axes are equal in length and $b$ becomes the radius $r$ of the circle, with the integral collapsing to $\pi / 2$, and we recover the circumference of a circle. When the eccentricity is $1$, $a$ is forced to be $0$ and the integral collapsis to $1$, so that the perimeter of the ellipse is two times the length of the line segment $[-b, b]$ (you alluded to this case in your question).