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| sin2φ=2y′/1+y′2; | cos2φ=1−y′2/1+y′2; |
| X= | x+vsin2φ=x−y′/y″, |
| Y= | y+1−y′2/2y″. |
So far all is general; in the particular example proposed,
y2=4ax; y=2a1/2x1/2, y′=a1/2x−1/2, y″=−1/2a1/2x−3/2;
| ∴ X | = | x+2x=3x, | ||
| Y | = | y+1−ax−1/−a1/2x−3/2 | = | 2a1/2x1/2−x3/2−ax1/2/a1/2 |
| = | 3ax1/2−x3/2/a1/2 | |||
| = | 9a−X/3√3a1/2X1/2. |
From this it appears that Y=0, or the curve crosses the axis, where X=9a, which answers to the point in the parabola for which x=3a,
| dy/dx | = | 1/3√3a1/2{9a−X/2X1/2−X1/2} |
| = | 1/6√3a1/2·9a−3X/X1/2. |
When x=0 this is infinite, so that the caustic like the reflecting curve is perpendicular to the axis at its origin: when
x=9a, Y=0; dY/dX=1/6√3a1/2·−18a/3a1/2=−1/√3.
The angle at which the caustic afterwards cuts the axis is therefore that having for its natural tangent 1/√3, which shows it to be one of 30°.
The curve extends without limit in the same directions with its generating parabola.
