Hence the orbit will be an Ellipse, Hyperbola, or Parabola, according as

It is consequently an Ellipse, and the equation becomes, since

Hence also, if T, T' denote the Periodic Times in this Ellipse, and in the circle whose radius is 2R the given altitude,

445.

Let APV (Fig. 93,) be the given cycloid, AB its

base, and VB its axis. Then the figure being completed in Newton's manner, as in the diagram, we have

RP QT :: ZP2: ZT2 :: VF2 : EF2
:: VB: BE

and since chord of curvature PV 4PM, therefore
RP2 4PM. RQ

.. 4PM × RQ : QT2 :: VB: PM

P, P' be the perpendiculars upon the tangent corresponding to the greatest and least velocities; then by the question, we have

or the point required is at the extremity of the minor axis.

P

√ a2 —b2). (a + √ a2 —b2)

Hence, if T, T' denote the times down the first last halves of the r, we have

Since F c

the curve is a conic section with

449. force in the focus; and since moreover, by the question, it turns into itself again, it must be either an ellipse or a circle. When the angle of reflection or angle of incidence a is 45°, and

SQ
2

QA (Fig. 94,) = then the orbit is a circle whose radius is SQ,

because then SQ is at right angles to the tangent Tt, and the velocity in a circle is that which is acquired down its radius. In all other

cases the orbit is an Ellipse, of which S is one focus.

Let QA r, SQR; then the perpendicular upon the tangent Tt, is

PR. sin. (— 2a) = R. sin, 2a

and by 441 the equation to the Ellipse is

But if V be the velocity acquired down r, then

Again, by 448 the whole time of descent to the centre is

and by 440, the Periodic Time in the circle, whose radius is SA, is 2 V (R+r)

μ (la) 2
g

2

r being that value of which corresponds to t = 0.

Hence the whole time to the centre is

and the Periodic Time in a circle whose radius is r and force =

451.

Let T = ma2 be the Periodic Time. Then it is to

T as area described in dt is to the whole Ellipse, or we have