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

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:
:400px

Proof:
:400px
Let the points be labelled to simplify:
:$A := \tuple {1, 4}$
:$B := \tuple {2, 8}$
:$C := \tuple {4, 2}$
:$D := \tuple {8, 5}$
:$E := \tuple {7, 1}$
:$F := \tuple {5, 7}$
Let $ABCDEF$ be considered as a hexagon.
We join the opposite points of $ABCDEF$:
:$AF: \tuple {1, 4} \to \tuple {5, 7}$
:$BC: \tuple {2, 8} \to \tuple {4, 2}$
:$BE: \tuple {2, 8} \to \tuple {7, 1}$
:$AD: \tuple {1, 4} \to \tuple {8, 5}$
:$CD: \tuple {4, 2} \to \tuple {8, 5}$
:$EF: \tuple {7, 1} \to \tuple {5, 7}$
It is to be shown that the intersections of:
:$AF$ and $BC$
:$BE$ and $AD$
:$CD$ and $EF$
all lie on the same straight line.
The result then follows from Pascal's Mystic Hexagram.
From Two-Point Form of Equation of Straight Line in Plane:
:$\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$
Thus:
{{begin-eqn}}
{{eqn | n = AF
      | l = \frac {y - 4} {x - 1}
      | r = \frac {7 - 4} {5 - 1}
      | c = 
}}
{{eqn | r = \frac 3 4
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 4 \paren {y - 4}
      | r = 3 \paren {x - 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac 3 4 x + \dfrac {13} 4
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = BC
      | l = \frac {y - 8} {x - 2}
      | r = \frac {2 - 8} {4 - 2}
      | c = 
}}
{{eqn | r = -3
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y - 8
      | r = -3 \paren {x - 2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = -3 x + 14
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = BE
      | l = \frac {y - 8} {x - 2}
      | r = \frac {1 - 8} {7 - 2}
      | c = 
}}
{{eqn | r = -\frac 7 5
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 5 \paren {y - 8}
      | r = -7 \paren {x - 2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 5 y - 40
      | r = -7 x + 14
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = -\frac 7 5 x + \frac {54} 5
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = AD
      | l = \frac {y - 4} {x - 1}
      | r = \frac {5 - 4} {8 - 1}
      | c = 
}}
{{eqn | r = \frac 1 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 7 \paren {y - 4}
      | r = x - 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 7 y - 28
      | r = x - 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \frac x 7 + \frac {27} 7
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = CD
      | l = \frac {y - 2} {x - 4}
      | r = \frac {5 - 2} {8 - 4}
      | c = 
}}
{{eqn | r = \frac 3 4
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 4 \paren {y - 2}
      | r = 3 \paren {x - 4}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 4 y - 8
      | r = 3 x - 12
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \frac 3 4 x - 1
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = EF
      | l = \frac {y - 1} {x - 7}
      | r = \frac {7 - 1} {5 - 7}
      | c = 
}}
{{eqn | r = -3
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y - 1
      | r = -3 x + 21
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = -3 x + 22
      | c = 
}}
{{end-eqn}}
Evaluate the intersection of $AF$ and $BC$:
{{begin-eqn}}
{{eqn | l = y
      | r = \dfrac 3 4 x + \dfrac {13} 4
      | c = 
}}
{{eqn | l = y
      | r = -3 x + 14
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac 3 4 x + \dfrac {13} 4
      | r = -3 x + 14
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 3 x + 13
      | r = -12 x + 56
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 15 x
      | r = 43
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \dfrac {43} {15}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = -3 \paren {\dfrac {43} {15} } + 14
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac {-129 + 210} {15}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac {81} {15}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac {27} 5
      | c = 
}}
{{end-eqn}}
So $AF$ and $BC$ intersect at $\paren {\dfrac {43} {15}, \dfrac {27} 5}$.
Evaluate the intersection of $BE$ and $AD$:
{{begin-eqn}}
{{eqn | l = y
      | r = -\frac 7 5 x + \frac {54} 5
      | c = 
}}
{{eqn | l = y
      | r = \frac x 7 + \frac {27} 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\frac 7 5 x + \frac {54} 5
      | r = \frac x 7 + \frac {27} 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -49 x + 378
      | r = 5 x + 135
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 54 x
      | r = 243
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \dfrac 9 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \frac 1 7 \paren {\frac 9 2} + \frac {27} 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac {9 + 54} {14}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac {63} {14}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac 9 2
      | c = 
}}
{{end-eqn}}
So $BE$ and $AD$ intersect at $\paren {\dfrac 9 2, \dfrac 9 2}$.
Evaluate the intersection of $CD$ and $EF$:
{{begin-eqn}}
{{eqn | l = y
      | r = \frac 3 4 x - 1
      | c = 
}}
{{eqn | l = y
      | r = -3 x + 22
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac 3 4 x - 1
      | r = -3 x + 22
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac 3 x - 4
      | r = -12 x + 88
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 15 x
      | r = 92
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \dfrac {92} {15}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = -3 \paren {\dfrac {92} {15} } + 22
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac {-92 + 110} 5
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac {18} 5
      | c = 
}}
{{end-eqn}}
So $CD$ and $EF$ intersect at $\paren {\dfrac {92} {15}, \dfrac {18} 5}$.
It remains to be shown that those points of intersection:
:$\paren {\dfrac {43} {15}, \dfrac {27} 5}$, $\paren {\dfrac 9 2, \dfrac 9 2}$, $\paren {\dfrac {92} {15}, \dfrac {18} 5}$
all lie on the same straight line.
From Two-Point Form of Equation of Straight Line in Plane:
:$\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$
Thus:
{{begin-eqn}}
{{eqn | l = \frac {y - \frac {27} 5} {x - \frac {43} {15} }
      | r = \frac {\frac {18} 5 - \frac {27} 5} {\frac {92} {15} - \frac {43} {15} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {5 y - 27} {15 x - 43}
      | r = \frac {18 - 27} {92 - 43}
      | c = simplifying 
}}
{{eqn | ll= \leadsto
      | l = \frac {5 y - 27} {15 x - 43}
      | r = -\frac 9 {49}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 245 y - 1323
      | r = - 135 x + 387
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 245 y
      | r = - 135 x + 1710
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 49 y + 27 x
      | r = 342
      | c = 
}}
{{end-eqn}}
It remains to demonstrate that $\paren {\dfrac 9 2, \dfrac 9 2}$ lies on this line:
{{begin-eqn}}
{{eqn | l = 49 \dfrac 9 2 + 27 \dfrac 9 2
      | r = \dfrac {441 + 243} 2
      | c = 
}}
{{eqn | r = \dfrac {684} 2
      | c = 
}}
{{eqn | r = 342
      | c = 
}}
{{end-eqn}}
Bingo.
{{qed}}