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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,

$tan PvN =-tan vPQ =-tan2 NPQ =-tan2 PNv$ .

And since $PN$ is a normal,

$tan PNv = dx / dy$ ; $∴ tan2 PNv = 2 dx / dy / 1- dx 2 / dy 2 = 2 dxdy / dy 2 - dx 2$ .

The equation is therefore

$Y - y +2 dxdy / dy 2 - dx 2 (X - x)=0$ (1);

and we have to put for $dxdy / dy 2 - dx 2$ 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

$Y - y +2\cdot tan θ / 1-tan θ 2 (X - x)=0$ .

Example. Suppose the curve to be a common parabola, its equation is $y 2 =4 αx$ ,

$tan θ = dx / dy = y / 2 α$ ;