Page:Optics.djvu/46

22

∴ y=2atanθ; x=y²/4a=atanθ².

Then if we put t for tanθ,

Y−2at+2t/1−t2(X−at2)	=0;
∴ Y(1−t2)+2(X−a)t	=0⁠(1).

Then differentiating with respect to t,

(2);

−ty+X−a=0; ∴ t=X−a/Y

and substituting this value of t in (1), we have

Y(1−(x−a)²/Y²)+2(x−a)²/Y, or Y²+(x−a)²=0;

the equation to a point, namely, the focus, where X=a.

27. The method pursued in the following example is perhaps less elegant than that just given, but it has often the advantage of being simpler and less prolix.

Required the form of the caustic, when the reflecting curve is a common parabola, and the incident rays are perpendicular to the axis.

Let P (Fig. 21.) be a point of the curve MP, Pq an incident and a reflected ray. Then, taking for granted that when q is a point of the caustic, Pq is one-fourth of the chord of the circle of curvature at P, we have the following easy method of determining the co-ordinates of q.

sinφ=y′/√1+y′2;

cosφ=1/√1+y′2;