Page:Optics.djvu/47

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sin2 φ = 2 y' / 1+ y' 2

;

cos2 φ = 1- y' 2 / 1+ y' 2

;

$X =$	$x + v sin2 φ = x - y' / y ″$ ,
$Y =$	$y + 1- y' 2 / 2 y ″$ .

So far all is general; in the particular example proposed,

$y 2 =4 ax$ ; $y =2 a 1 / 2 x 1 / 2$ , $y'= a 1 / 2 x - 1 / 2$ , $y ″=- 1 / 2 a 1 / 2 x - 3 / 2$ ;

$∴ X$	$=$	$x +2 x =3 x$ ,
$Y$	$=$	$y + 1- ax -1 / - a 1 / 2 x - 3 / 2$	$=$	$2 a 1 / 2 x 1 / 2 - x 3 / 2 - ax 1 / 2 / a 1 / 2$
			$=$	$3 ax 1 / 2 - x 3 / 2 / a 1 / 2$
			$=$	$9 a - X / 3 \sqrt 3 a 1 / 2 X 1 / 2$ .

From this it appears that $Y =0$ , or the curve crosses the axis, where $X =9 a$ , which answers to the point in the parabola for which $x =3 a$ ,

$dy / dx$	$=$	$1 / 3 \sqrt 3 a 1 / 2 {9 a - X / 2 X 1 / 2 - X 1 / 2$ }
	$=$	$1 / 6 \sqrt 3 a 1 / 2 \cdot 9 a -3 X / X 1 / 2$ .

When $x =0$ this is infinite, so that the caustic like the reflecting curve is perpendicular to the axis at its origin: when

$x =9 a$ , $Y =0$ ; $dY / dX = 1 / 6 \sqrt 3 a 1 / 2 \cdot -18 a / 3 a 1 / 2 = -1 / \sqrt 3$ .

The angle at which the caustic afterwards cuts the axis is therefore that having for its natural tangent $1 / \sqrt 3$ , which shows it to be one of $30°$ .

The curve extends without limit in the same directions with its generating parabola.

Page:Optics.djvu/47

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