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sin2φ=2y′1+y′2; | cos2φ=1−y′21+y′2; |
X= | x+vsin2φ=x−y′y″, |
Y= | y+1−y′22y″. |
So far all is general; in the particular example proposed,
y2=4ax; y=2a12x12, y′=a12x−12, y″=−12a12x−32;
∴ X | = | x+2x=3x, | ||
Y | = | y+1−ax−1−a12x−32 | = | 2a12x12−x32−ax12a12 |
= | 3ax12−x32a12 | |||
= | 9a−X3√3a12X12. |
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,
dydx | = | 13√3a12{9a−X2X12−X12} |
= | 16√3a12·9a−3XX12. |
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; dYdX=16√3a12·−18a3a12=−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.