Download 51. A Parabola as an Envelope

51. A Parabola as an Envelope
On one side of an arbitrary angle, mark off n equally spaced points 1, 2, . . . , n from the
vertex (0), and on the other side n equally spaced points (not necessarily with the same
spacing on the first side) n " 1, n " 2, . . . , 2, 1, 0 from the vertex (n). Prove that the lines
joining points with the same number on both sides envelope (i.e. are tangent to) a parabola.
A=12
1
2
3
4
7
6
5
9
8
10
11
S
12 11
4
3
2
B=1
Figure 1
The proof uses the
Theorem of Apollonius. If two parabola tangents SA and SB, with the points of tangency
A and B intersect a third parabola tangent at P and Q with O the point of tangency on
SP
QO
QS
.
the parabola, then PA
OP
SQ
A
P
S
O
Q
B
Figure 2
The proof of this theorem follows from the known property of parabolas that the point of
intersection of two parabola tangents lies on a line parallel to the parabola’s
axis, and passing through the midpoint of the chord joining the points of
tangency. (This is easily established from the equation y 4p1 x 2 of the parabola
with focus FŸ0, p , directrix y "p and axis x 0. )
It follows from this theorem that if A and B are two points on a parabola, A and B their
U
U
1
projections on the directrix, M the midpoint of segment AB, and M the projection of
M on the directrix, then M is the midpoint of A B , i.e., AM M B.
U
U
U
U
U
U
A
M
F
A'
B
M'
d ire c trix
B'
Figure 3
Proof of the Theorem of Apollonius Let AP a, PS ), BQ b, QS *, OP p and
OQ q. Let a , ) , b , . . . be the projections of AP, PS, BQ, . . . on the directrix. Then
U
U
U
(1)
p
U
(2)
q
U
(3)
b *
U
(4)
) *
U
U
U
a
U
b
U
a )
U
p q
U
U
U
From (1), (2) and (4), we get ) * a b , and then from (3), ) b and * a .
Then
U
U
U
U
U
SP
PA
)
a
)
aU
bU
aU
,
QO
OP
q
p
qU
pU
bU
aU
,
BQ
QS
*
b
bU
bU
aU
,
which proves the Theorem of Apollonius.
U
*
U
U
U
U
R
Proof of the envelope. Let S be the vertex of the given angle, e and f the length of the
segments marked off n times on each side of the angle, and A and B those points of
the sides so that SA ne and SB nf, with A numbered n and B numbered 0.
(Figure 1 shows the case n 12. ) Consider the parabola that is tangent to SA and
SB at A and B. (The reader can use analytic geometry, for instance, to show that a
unique such parabola exists.) The tangent line joining the point P numbered 7 on SA
SP
BQ
meets the other side in a point Q so that PA
by Apollonius’ Theorem. But
QS
SP
v
n"v , and it follows that Q is the point onf SB numbered 7. Thus all the lines
PA
R
drawn connecting like numbered points are tangent to a parabola.
Note 1. This is the idea behind a lot of "string art".
Note 2. It’s a good exercise to use some sort of geometry software program to
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implement this "construction".
Note 3. One can use Apollonius’ Theorem to construct the point of tangency on each
QO
SP
line. (From Figure 2, OP PA
n"v v . )
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