three theorems on parabolas

three theorems on parabolas

In the

Cartesian plane

, pick a point with

coordinates

(

0

,

2

⁢

f

)

(subtle hint!) and construct (1) the set

S

of

segments

s

joining

F

=

(

0

,

2

⁢

f

)

with the
points

(

x

,

0

)

, and (2) the set

B

of right-bisectors

b

of the segments

s

∈

S

.

Theorem 1 :

The

envelope

described by the lines of the set

B

is a

parabola

with

x

-axis as

directrix

and focal length

|

f

|

.

Proof:

We’re lucky in that we don’t need a fancy

definition

of envelope; considering a line to be a set of points it’s just the boundary of the set

C

=

∪

b

∈

B

b

. Strategy: fix an

x

coordinate and find the max/

minimum

of possible

y

’s in C with that

x

. But first we’ll pick an

s

from

S

by picking a point

p

=

(

w

,

0

)

on the

x

axis. The

midpoint

of the segment

s

∈

S

through

p

is

M

=

(

w

2

,

f

)

. Also, the slope of this

s

is

-

2

⁢

f

w

. The corresponding right-bisector will also

pass through

(

w

2

,

f

)

and will have slope

w

2

⁢

f

. Its equation is therefore

2

⁢

y

-

2

⁢

f

2

⁢

x

-

w

=

w

2

⁢

f

.

Equivalently,

y

=

f

+

w

⁢

x

2

⁢

f

-

w

2

4

⁢

f

.

By any of many very famous

theorems

(Euclid book II theorem twenty-something, Cauchy-Schwarz-Bunyakovski (overkill),

differential calculus

, what
you will) for fixed

x

,

y

is an

extremum

for

w

=

x

only, and therefore the envelope has equation

y

=

f

+

x

2

4

⁢

f

.

I could say I’m done right now because we “know” that this is a parabola, with focal length

f

and

x

-axis as directrix. I don’t want to, though. The
most popular definition of parabola I know of is “set of points equidistant from some line

d

and some point

f

.” The line responsible for the point on
the envelope with given

ordinate

x

was found to bisect the segment

s

∈

S

through

H

=

(

x

,

0

)

. So pick an extra point

Q

∈

b

∈

B

where

b

is
the

perpendicular bisector

of

s

. We then have

∠

⁢

F

⁢

M

⁢

Q

=

∠

⁢

Q

⁢

M

⁢

H

because they’re both

right angles

, lengths

F

⁢

M

=

M

⁢

H

, and

Q

⁢

M

is
common to both

triangles

F

⁢

M

⁢

Q

and

H

⁢

M

⁢

Q

. Therefore two sides and the angles they contain are respectively equal in the triangles

F

⁢

M

⁢

Q

and

H

⁢

M

⁢

Q

, and so respective angles and respective sides are all equal. In particular,

F

⁢

Q

=

Q

⁢

H

. Also, since

Q

and

H

have the same

x

coordinate, the line

Q

⁢

H

is the

perpendicular

to the

x

-axis, and so

Q

, a general point on the envelope, is equidistant from

F

and the

x

-axis.
Therefore etc.

QED.

Because of this construction, it is clear that the lines of

B

are all

tangent

to the parabola in question.

We’re not done yet. Pick a random point

P

outside

C

(“inside” the parabola), and call the parabola

π

(just to be nasty). Here’s a nice quicky:

Theorem 2 The Reflector Law:

For

R

∈

π

, the length of the path

P

⁢

R

⁢

F

is

minimal

when

P

⁢

R

produced is perpendicular to the

x

-axis.

Proof:

Quite simply, assume

P

⁢

R

produced is not necessarily perpendicular to the

x

-axis. Because

π

is

a parabola, the segment from

R

perpendicular to the

x

-axis has the same length as

R

⁢

F

. So let this perpendicular hit the

x

-axis at

H

. We then have that the length of

P

⁢

R

⁢

H

equals that of

P

⁢

R

⁢

F

. But

P

⁢

R

⁢

H

(and hence

P

⁢

R

⁢

F

) is minimal when it’s a

straight line

; that is, when

P

⁢

R

produced is perpendicular to the

x

-axis.

QED

Hey! I called that theorem the “reflector law”. Perhaps it didn’t look like one. (It

is

in the Lagrangian formulation), but it’s fairly easy to show (it’s a

similar

argument

) that the shortest path from a point to a line to a point makes and “reflected” angles equal.

One last marvelous tidbit. This will take more time, though. Let

b

be tangent to

π

at

R

, and let

n

be perpendicular to

b

at

R

. We will call

n

the

to

π

at

R

. Let

n

meet the

x

-axis at

G

.

Theorem 3 :

The radius of the “best-fit circle” to

π

at

R

is twice the length

R

⁢

G

.

Proof:

(Note: the

≈

’s need to be phrased in terms of upper and lower bounds, so I can use the

sandwich theorem

, but the proof schema is exactly what is
required).

Take two points

R

,

R

′

on

π

some small

distance

ϵ

from each other (we don’t actually use

ϵ

, it’s just a psychological trick).
Construct the tangent

t

and normal

n

at R, normal

n

′

at

R

′

. Let

n

,

n

′

intersect

at

O

, and

t

intersect the

x

-axis at

G

.

R

⁢

F

,

R

′

⁢

F

.

Erect perpendiculars

g

,

g

′

to the

x

-axis through

R

,

R

′

respectively.

R

⁢

R

′

. Let

g

intersect the

x

-axis at

H

. Let

P

,

P

′

be points on

g

,

g

′

not in

C

. Construct

R

⁢

E

perpendicular to

R

⁢

F

with

E

in

R

′

⁢

F

. We now have

i)

∠

⁢

P

⁢

R

⁢

O

=

∠

⁢

O

⁢

R

⁢

F

=

∠

⁢

G

⁢

R

⁢

H

≈

∠

⁢

P

′

⁢

R

′

⁢

O

=

∠

⁢

O

⁢

R

′

⁢

F

iii)

E

⁢

R

≈

F

⁢

R

⋅

∠

⁢

E

⁢

F

⁢

R

v)

∠

⁢

R

′

⁢

R

⁢

E

+

∠

⁢

E

⁢

R

⁢

O

≈

π

2

(That’s the number

π

, not the parabola)

vii)

∠

⁢

E

⁢

R

⁢

O

+

∠

⁢

O

⁢

R

⁢

F

=

π

2

ix)

∠

⁢

R

′

⁢

E

⁢

R

≈

π

2

xi)

∠

⁢

R

′

⁢

O

⁢

R

=

1

2

⁢

∠

⁢

R

′

⁢

F

⁢

R

xiii)

R

′

⁢

R

≈

O

⁢

R

⋅

∠

⁢

R

′

⁢

O

⁢

R

xv)

F

⁢

R

=

R

⁢

H

From (iii),(iv) and (i) we have

∠

⁢

R

′

⁢

R

⁢

E

≈

∠

⁢

G

⁢

R

⁢

H

, and since

R

′

is close to

R

, and if we let

R

′

approach

R

, the approximations
approach equality. Therefore, we have that triangle

R

′

⁢

R

⁢

E

approaches

similarity

with

G

⁢

R

⁢

H

. Therefore we have

R

⁢

R

′

:

E

⁢

R

≈

R

⁢

G

:

R

⁢

H

.
Combining this with (ii),(vi),(vii), and (viii) it follows that

R

⁢

O

≈

2

⁢

R

⁢

G

, and in the limit

R

′

→

R

,

R

⁢

O

=

2

⁢

R

⁢

G

.

QED

This last theorem is a very nice way of short-cutting all the messy

calculus

needed to derive the Schwarzschild “Black-Hole” solution to Einstein’s equations, and that’s why I enjoy it so.

Title

three theorems on parabolas

Canonical

name

ThreeTheoremsOnParabolas

Date of creation

2013-03-22 12:40:51

Last modified on

2013-03-22 12:40:51

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

19

Author

CWoo (3771)

Entry type

Topic

Classification

msc 51N20

Related topic

PropertiesOfParabola

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