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myarbrough@oreilly.com 2014-11-18 07:13:15 -08:00
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@ -157,7 +157,7 @@ There is also((("+ operator")))((("elliptic curve cryptography","addition operat
Geometrically, this third point P~3~ is calculated by drawing a line between P~1~ and P~2~. This line will intersect the elliptic curve in exactly one additional place. Call this point P~3~' = (x, y). Then reflect in the x-axis to get P~3~ = (x, –y). Geometrically, this third point P~3~ is calculated by drawing a line between P~1~ and P~2~. This line will intersect the elliptic curve in exactly one additional place. Call this point P~3~' = (x, y). Then reflect in the x-axis to get P~3~ = (x, –y).
There are a couple of special cases which explain the need for the "point at infinity." There are a couple of special cases that explain the need for the "point at infinity."
If P~1~ and P~2~ are the same point, the line "between" P~1~ and P~2~ should extend to be the tangent on the curve at this point P~1~. This tangent will intersect the curve in exactly one new point. You can use techniques from calculus to determine the slope of the tangent line. These techniques curiously work even though we are restricting our interest to points on the curve with two integer coordinates! If P~1~ and P~2~ are the same point, the line "between" P~1~ and P~2~ should extend to be the tangent on the curve at this point P~1~. This tangent will intersect the curve in exactly one new point. You can use techniques from calculus to determine the slope of the tangent line. These techniques curiously work even though we are restricting our interest to points on the curve with two integer coordinates!