Harriot, Thomas - Mss. 6787

748375

4.) De parabola
5. casuum:

[Translation: 4) On the parabola.
Case 5; ]

Sunt alij
casus.
Vide dorsum
chart: b.)
de
[Translation: There are other cases. See the back of sheet b) on the ]

Cum duplex
sit

b s

, duæ
erunt para-
polæ et ver-
tex alterius
erit inter

D

a

[Translation: Since there are two cases for

b s

, there are two parabolas and the vertex of the other will be between

D

and

a

problema.
Datis tribus punctis,
quorum duo sunt
in parabola, et
tertium in centroide:
Invenire
[Translation: Problem.
Given three points of which two are on the parabola and the third is at the focus, find the ]

Sint tria data puncta

a

b

c

. Sint

b

c

, in
parabola, et

a

in
centroide.
Connectantur: et circa
tres lineas ut diametros,
fiunt tres circuli. Quorum duo, circa

a b

b c

semivicem secant
in punctis

b

t

, quæ inugantur et fit

b t

perpendicularis

a c

.
Deinde centro

a

. Intervallo

a c

, fiat circulus

c f

. Itidem centro

a

, intervallo

a b

fiat circulum

u b d

, qui secabit

a c

u

.
Unde

u c

est differentia inter

a b

a c

:
Fiat

b s = u c

. Agatur

c s

usque ad

k

. Agatur

k a

cum proeluctione
quae secabit circulum (circa

a

centrum), in

d

f

: et
circa

a b

diametrum, in

h

. Agatur

b h

quæ erit parallela et
aequalis

s k

. Ita

h k

b s

. Bisecetur

h d

in puncto

g

g

etiam est medium punctum inter

k

f

.

[Translation: Let the three given points be

a

b

c

. Let

b

and

c

be on the parabola and

a

the focus.
They are connected and around the three lines as diameters are created three circles. Of which two, around

a b

and

b c

in turn cut in the points

b

t

, which are joined and

b t

is perpendicular to

a c

.
Then with centre

a

, radius

a c

, create a circle

u b d

, which cuts

a c

u

.
Whence

u c

is the difference between

a b

and

a c

.
Let

b s = u c

. Take

c s

as far as

k

. Take

k a

with its extension which will cut the circle with centre

a

d

and

f

; and the circle on the diameter

a b

h

. Take

b h

which will be parallel and equal to

s k

. Thus

h k

and

b s

h d

is bisected at the point

g

g

is therefore the midpoint between

k

and

f

Dico quod; punctum

g

est vertex

g k

axis diameter; et

h b m

k l m

lineæ ordinatim appli-

[Translation: I say that the point

g

is the vertex of the parabola:

g k

is the diametric axis; and

h b m

k l m

, ordinates.

Ad exegesin arithmeticam agatur

t s

. et

s y

perpendicularis

t c

b t s c

est quadrilaterum in circulo. Dantur omnes lineæ et per-
pendicularis

s y

. Tum

c s : s y : c a : a k

. Deinde

\frac{a c + a k}{2} = k g

[Translation: To show the arithmetic take

t s

, and

s y

perpendicular to

t c

b t s c

is a cyclic quadrilateral. There are given all the lines and perpendiculars

s y

. Then

c s

s y

c a

a k

are proportionals. Thence

\frac{a c + a k}{2} = k g

Problema igitur satis
[Translation: The problem is therefore satisfactorily ]

691 (346v)	692 (347)
693 (347v)	694 (348)
695 (348v)	696 (349)
697 (349v)	698 (350)
699 (350v)	700 (351)

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