Then from (1) we have

(a + a)2 + (b + b)o _ a2 + b2 = (1 + 1)", " + ".

From (2) we have

B)

a2

2

(a + a) (b + B) = (1 + 1) ab

Neglecting squares and products of the small quantities a, 6, 1,

and putting cos = 1, sin = 4, this becomes

2

1

a2 + b2 = 1 (1 - 2a) (cos° 0 – 24. cos ◊ sin 0)

b*

a2

17. If a point describe a parabola, the acceleration being towards the focus; shew that the time of describing any arc bounded by a focal chord (length of chord).

1

1+ cos 0, λ=?

And the time required = { ["__»o d0, where ASQ = ß,

В-п

1 + cos ẞ 1+ cos (B-T) sin3ß

therefore the time ∞ (chord).

A. 18. A number of points move in hyperbolas, starting from their vertices at the same instant; the directions of the accelerations pass through the common center, and their magnitudes are equal at all equal distances. Shew that if the major axes coincide, the points will always lie in a common ordinate: and if the asymptotes coincide, they will always lie in a straight line through the

center.

From Art. 70, note, we have, for any one of the moving points,

At any assigned instant of time, let x,y,, x,y,, &c. be the coordinates of the points; a,b,, a,b,, &c. being the elements of their paths.

A. 19. To find the law of acceleration towards the node of a lemniscate, in order that a point may move in that curve.

d2

... a = h3u2 (u + du) = зa'h3u”, which ∞ (distance)-'.

A. 20. The acceleration towards a fixed point is at distancer: a point is initially moving with velocity direction at right angles to its initial distance (a) from point. Find the orbit described.

with a velocity 2a√3A at right angles to its initial distance a: shew that it will come to a second apse at distance 3a.

the first of which gives the original apsidal distance, the other gives an apsidal distance 3a.

* In such cases as this, the acceleration is supposed to be towards the origin unless the contrary be distinctly expressed.

L.