Is it possible to find the distance between two points on the circumference of an ellipse following the outer curve? I am rather confused on this calculation, and I can't seem to find a solution online. I need to know how to find the distance between two known points on an ellipse's outer edge following the path of the curve. For example:
Example
In that image, how would I go about using the coordinates of point A and B to find the arc length/segment of the circumference between points A and B. The current coordinates for A in this picture is (0.4, 0.98) and B is (1.4, 0.72), and the ellipse's equation is (x^2/2^2)+(y^2/1^2)=1. Thank you!
 A: The arc length from $x=a$ to $x=b$ on the top half of the ellipse is
$$ \int_a^b \sqrt{\frac{16 - 3x^2}{16 - 4 x^2}}\; dx $$
The antiderivative is not elementary, but can be expressed in terms of incomplete elliptic integrals: (in Maple's notation)
$$ 2 {\it EllipticE} \left(\frac{b}{2}, \frac{\sqrt{3}}{2}\right) - 2 
 {\it EllipticE} \left(\frac{a}{2}, \frac{\sqrt{3}}{2}\right)$$
Alternatively, you can use numerical integration.
In the case at hand, with $a = 0.4$ and $b = 1.4$, the result is  approximately $1.03990644$.
A: 
By Pythagoras theorem you have
$$\text{$\Delta $L}=\sqrt{\text{$\Delta $x}^2+\text{$\Delta $y}^2}$$
Collect $\Delta x$ to get
$$ \text{$\Delta $L}=\sqrt{\text{$\Delta $x}^2 \left(\frac{\text{$\Delta $y}^2}{\text{$\Delta $x}^2}+1\right)}=\text{$\Delta $L}=\text{$\Delta $x} \sqrt{1+\left(\frac{\Delta y}{\Delta x}\right)^2}$$
Now let $\Delta x\to 0$. If the function is differentiable you get
$${dL}= \sqrt{\left(1+\frac{{dy}}{{dx}}\right)^2}\;{dx}=\sqrt{\left(1+y'\right)^2}\;dx$$
On the interval $[a,b]$ you get
$$L=\int_a^b \sqrt{1+\left(y'\right)^2} \, dx$$
The ellipse on the halfplane  where $y>0$ has equation $y=\frac{\sqrt{4-x^2}}{2}$ and its derivative is $y'=-\frac{x}{2 \sqrt{4-x^2}}$ therefore you have
$$L=\int_{0.4}^{1.4} \sqrt{1+\left(-\frac{x}{2 \sqrt{4-x^2}}\right)^2} \, dx=\frac{1}{2} \int_{0.4}^{1.4} \sqrt{\frac{3 x^2-16}{x^2-4}} \, dx$$
This integral is about $L\approx 1.04$ and cannot be calculated with usual functions. That's why the other posters called it "elliptic".
For small intervals, however, it can be approximated by
$$L\approx h \left(1+\frac{1}{2} \left(\frac{b x_0}{a^2}\right)^2\right)$$
where $a$ and $b$ are the parameters of the ellipse, $x_0$ the starting point and $h$ the interval $x_B-x_A$
In your example it gives $L\approx 1.005$ which is a poor approximation, but if you consider $A$ and $B$ closer, let's say $h=0.25$ we get $L\approx 0.251$ while the "exact" formula gives $L\approx 0.252$ which is a fair approximation.
I hope this helps. I preferred a more elementary approach because my intuition suggested you don't eat elliptic integrals for breakfast :)
A: I think that parametrisation could make things a bit easier. 
Consider the general right ellipse equation $$\left(\frac x a \right)^2+\left(\frac yb \right)^2=1$$ and use $$x=a \cos(t)\qquad y=b \sin(t)$$ This makes the arc length to be $$I=\int_{t_1}^{t_2} \sqrt {x'^2+y'^2}\,dt=\int_{t_1}^{t_2} \sqrt {a^2 \sin ^2(t)+b^2 \cos ^2(t)}\,dt$$ Integrating, this leads for the antiderivative to $$\int\sqrt {a^2 \sin ^2(t)+b^2 \cos ^2(t)}\,dt=b E\left(t\left|1-\frac{a^2}{b^2}\right.\right)$$ where appears the elliptic integral of the second kind; then $$I=b \left(E\left(t_2\left|1-\frac{a^2}{b^2}\right.\right)-E\left(t_1\left|1-\frac{a^2}{b^2}\right.\right) \right)$$
