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principal co-ordinate axis. Let AN (Fig. 20.) be this axis, AM, MP co-ordinates of the curve, PN a normal, QP, Pv an incident and a reflected ray. The question is first to determine the equation of the line Pv. Since this line passes through the point P whose co-ordinates are x, y, the equation must be

Y−y=α(X−x),

α being the tangent of the angle PvN.

Now,

tanPvN=−tanvPQ=−tan2NPQ=−tan2PNv.

And since PN is a normal,

tanPNv=dx/dy; ∴ tan2PNv=2dx/dy/1−dx²/dy²=2dxdy/dy²−dx².

The equation is therefore

Y−y+2dxdy/dy²−dx²(X−x)=0⁠(1);

and we have to put for dxdy/dy²–dx² its value in terms of the co-ordinates given by the equation to the curve, and eliminate x and y between this, the equation (1), and its derivative.

26. The process is sometimes facilitated by taking for the variable a function of the angle PNM, as its tangent which is equal to dx/dy. The quantity we have called α is the tangent of twice this angle, and if we put θ for this angle, the equation to the reflected ray is

Y−y+2·tanθ/1−tanθ²(X−x)=0.

Example. Suppose the curve to be a common parabola, its equation is y²=4αx,

tanθ=dx/dy=y/2α;