Parabolic constant

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In mathematics, the ratio of the arc length of the parabolic segment formed by the latus rectum of any parabola to its focal parameter is a constant, denoted $\!\ P_2$ .

In the diagram, the latus rectum is pictured in blue, the parabolic segment that it forms in red and the focal parameter in green.

The value of $\!\ P_2$ is $\ln(1 + \sqrt2) + \sqrt2 \approx 2.29558714939...$ .

The circle and parabola are unique among conic sections in that they have a universal constant. The analogous ratios for ellipses and hyperbolas depend on their eccentricities. This means that all circles and parabolas are similar, and that all ellipses and hyperbolas are not.

1 Derivation
2 Properties
3 Applications
4 External links

[edit] Derivation

Take $y = \frac{x^2}{4a}$ as the equation of the parabola. The focal parameter is $p = 2 a$ and the semilatus rectum is $l = 2 a$ .

$\begin{align} P_2 := &\, \frac{1}{p}\int_{-l}^{l}\sqrt{1+\frac{dy^2}{dx^2}}\, dx \\ = &\, \frac{1}{2a}\int_{-2a}^{2a}\sqrt{1+\frac{x^2}{4a^2}}\, dx \\ = &\, \int_{-1}^{1}\sqrt{1+t^2}\, dt \quad (x=2at) \\ = &\, \operatorname{arsinh}(1)+\sqrt{2}\\ = &\, \ln(1+\sqrt{2})+\sqrt{2} \\ \end{align}$

[edit] Properties

The Lindemann-Weierstrass theorem easily shows that $\!\ P_2$ is transcendental. A proof by contradiction:

Suppose that $\!\ P_2$ is algebraic. If this is true, then $\!\ P_2 - \sqrt2 = \ln(1 + \sqrt2)$ must also be algebraic. However, by the Lindemann-Weierstrass theorem, $\!\ e^{\ln(1+ \sqrt2)} = 1 + \sqrt2$ would be transcendental, which is an obvious contradiction.