Points Defined by Adjacent Pairs of Digits of Reciprocal of 7 lie on Ellipse

Theorem
Consider the digits that form the recurring part of the reciprocal of $7$:
 * $\dfrac 1 7 = 0 \cdotp \dot 14285 \dot 7$

Take the digits in ordered pairs, and treat them as coordinates of a Cartesian plane.

It will be found that they all lie on an ellipse:


 * EllipseFromSeventh.png

Proof

 * EllipseFromSeventhSolution.png

Let the points be labelled to simplify:
 * $A := \left({1, 4}\right)$
 * $B := \left({2, 8}\right)$
 * $C := \left({4, 2}\right)$
 * $D := \left({8, 5}\right)$
 * $E := \left({7, 1}\right)$
 * $F := \left({5, 7}\right)$

Let $ABCDEF$ be considered as a hexagon.

We join the opposite points of $ABCDEF$:
 * $AF: \left({1, 4}\right) \to \left({5, 7}\right)$
 * $BC: \left({2, 8}\right) \to \left({4, 2}\right)$
 * $BE: \left({2, 8}\right) \to \left({7, 1}\right)$
 * $AD: \left({1, 4}\right) \to \left({8, 5}\right)$
 * $CD: \left({4, 2}\right) \to \left({8, 5}\right)$
 * $EF: \left({7, 1}\right) \to \left({5, 7}\right)$

From Equation of Straight Line through Two Points:


 * $\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$

Thus:

Evaluate the intersection of $AF$ and $BC$: