Natural Curve passing through three points I want to find the formula to a curve which passes through three points in $2D$. There is no other constraint and the more natural it looks the better, possibly zero tangent derivatives in $p_0$ and $p_2$. It doesn't matter which curve type (quadratic, cubic, etc) we use and the simplest solution that works is the best.

There are optional properties that would be desirable for a solution:


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*Being able to determine the length of the curve.

*Determine the $t$ variable at $p_1$.

*Determine tangents at $p_1$


The points can be colinear.
 A: is there anything wrong with the unique parabola going through those points? I'll assume you mean three points in general position (no two are colinear.)
In general, you can take $ax^2+bx+c=y$, and use all three points to determine a $3 \times 3$ system of linear equations to solve for coefficients.
In a similar fashion, you could take $(x-a)^2+(y-b)^2=r^2$ and again plug in $x,y$ to get a system of linear equations to solve.


*

*Both of these should go well, since tangent vectors amount to differentiating second order polynomials, which is easy.

*The parabola admits the easiest parametrization by $t$ given by $t \mapsto (t,t^2)$, while the circle can be given by $(x,y)=(a,b)+r(\cos t,\sin t)$ which is also not bad to evaluate.

*Arc lengths will be $\int_{t_1}^{t_2} \sqrt{x^{\prime}(t)^2+y^{\prime}(t)^2} dt$
which for the parabola is $\int_{t_1}^{t_2} \sqrt{1+4t^2} dt$ and for the circle $\int_{t_1}^{t_2} \sqrt{\cos^2t+\sin^2t} dt=\int_{t_1}^{t_2} 1 dt$. Maybe the only caveat is that if you want to find the arc length between points you have to find the appropriate $t$ that maps to a point $(x,y)$ but this is not so bad.
