■HMpp
[383
rr . ¿ _ Ti
r jnir>T»îilWIllliitiflÉHtyr P ftllTfT ^ 1 'ikVVWtrili
383]
that is, if
PROBLEMS AND SOLUTIONS.
A, +
+ +
V +
P
= 0,
A#! +
yLiOto +
va 3 +
p<* 4
= 0,
A/?i +
gß-i +
vß?, +
pßi
= 0,
A (c x + Yj)
+ + (Ca + Y2) +
V (c 3 + y 3 )
+ p (pi + Y4)
= 0,
ACiYi +
yc. 2 y, +
vc 3 Ys +
pCi Y4
= 0 :
607
and eliminating from these equations (A, ¡i, v, p), we find the above-mentioned relation
between a lt ft 1} y 1} c 1} &c.; which proves the theorem.
Ci 7i,
0,
the
[Vol. iv. pp. 70, 71.]
1771. (Proposed by Professor Cayley.)—Given a circle and a line, it is required
to find a parabola, having the line for its directrix, and the circle for a circle of
curvature.
2. Solution by the Proposer.
Let a? + y 2 — 1 = 0 be the equation of the given circle, x = m that of the given
line. Taking on the circle an arbitrary point (cos 0, sin 6), we may find a parabola
having the given line for its directrix, and touching the circle at the last-mentioned
point; viz., the equation of the parabola is found to be
y 2 — 2y sin 9 (1 + 2 cos 2 9 — 2m cos 9) — 4# cos 2 9 (cos 9 — m) + 1+3 cos 2 9 — 4m cos 9=0.
{There is no difficulty in verifying that this parabola has for its directrix the line
x — m = 0, that the equation is satisfied by the values x = cos 9, y = sin 9, and that the
derived equation is satisfied by the values x = cos 9, y = sin 9, ^ = — cot 9.)
Representing for a moment the left-hand side of the equation by U, we have identically
U — cos 2 9 (x 2 + y 2 — 1)
= y 2 sin 2 9 — a? cos 2 9 — 2y sin 9 (1 + 2 cos 2 9 — 2m cos 9) — 4x cos 2 9 (cos 9 — m)
+ 1 + 4 cos 2 9 — 4m cos 9,
= (;y sin 9 + x cos 9 — 1) {y sin 9 — x cos 9 — 1 — 4 cos 2 9 + 4m cos 9).
Hence to find the intersections of the parabola with the circle, we have first
x 2 + y 2 — 1 = 0, y sin 9 + x cos $ — 1=0,
giving the point (cos 9, sin 9) twice, since y sin 9 + x cos 0—1 = 0 is the equation of the
tangent to the circle at the point in question; and secondly
x 2 + y 2 — 1 = 0, y sin 9 — x cos 0—1—4 cos 2 0 + 4m cos 0 = 0,
