top of page
All
Articles

The Slope of an Elliptic Curve

Before we continue with our investigation of elliptic curves in a finite field, we need to clarify some important calculations, that will be used for adding and doubling values in a finite field, on an elliptic curve.


Finding the gradient (slope) of the tangent line at a point on an elliptic curve given in Weierstrass form involves using partial derivatives. The standard Weierstrass form of an elliptic curve is: y^2 = x^3 + Ax + B where A, B are constants in F , the field across which E, the elliptic curve, is defined.


To find the gradient at a point on this curve, you’ll need to differentiate implicitly since the equation involves both x and y interdependently.


Implicit Function - given the equation:

Differentiate implicitly with respect to x

Differentiate both sides of the equation with respect to x. Using implicit differentiation, the derivative of y^2 with respect to x is:

(applying the chain rule) and the derivative of:

with respect to x is:

Setting the derivatives equal gives:

To find dy/dx (the gradient of the tangent line at any point on the curve), rearrange the equation:

This equation gives you the slope of the tangent line at any point (x, y) on an elliptic curve in the Weierstrass form, provided that:

since dividing by zero is undefined.


Note that, in the form of the derivative where only x is used, the equation yields two solutions, precisely because there are 2 points, each reflected across the x-axis, for any value of x on the curve, each with a gradient g_n, such that g_1 = −g_2.

Comments


Green Juices
bottom of page