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PROBLEM II.

In an Ilyperbola, to find any Two Conjugate Diameters, an Ordinate to one of them, and its Absciss, one from another ; viz. baving any three of them

given, to find the Fourth.

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To find the Ordinate.
As
any

diameter :
Is to irs conjugate ::
So is the mean propor. between the abscisses :
To the ordinare.*

That is, as d :i:: ✓ id + *) * : Y iv id + x) x; where d denotes the diameter, c its conjugate, y an ordinate to the diameter, and x its abfcifs, or its distance from the vertex of the diameter, measured

upon

it. Note. In the hyperbola, the less abfciss added to the diameter, gives the greater absciss.

E X A M P L E.

If the diaineter be 24, its conjugate 21, and the less absciss 8 ; what is the ordinate? (d+- *) *

8 By the rule,

21/32 *

7 14 the ordinate required.

7 x 16

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d

24

8

CASE

* The values of the several quantities, in the four cases of this problem, are easily found from the general property of the curve, viz. dd :cc:: (+1)x : yy. The demonstration of which, together with that of the preceding problem, belongs to conic Sections,

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To find the Abscisses. As the conjugate: Is to the diameter :: So is the root of the sum of the squares of the

ordinate and semi-conjugate : To the distance between the foot of the ordinate

and the center. Then, to and from this distance, add and subtract the semi-diameter, and the sum and difference will give the two abscisles.

That is,
d/cctyy

+ d = x the greater or less absciss, according as the upper or under sign is used.

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The diameter and its conjugate being 24 and 21, required the two abscisses to the ordinate 14.

21

Here dvocc + yy

2475 X 212 + 142 +=

+ 12 4 W212 + 282

+ 12 = 4V 32 + 4 + 12 = 20 7 #12 = 32 and 8, the two abscisses required.

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To find the Conjugate.
As the mean propor. between the abscisses :
Is to the ordinate ::.
So is the diameter :
To its conjugate.

dy That is, id + *) * : y :: 0 :

(+ *)* Dd 2

EX

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What is the conjugate to the diameter 24, the less absciss being 8, and its ordinate 147

Here dy

24 X 14

= v (d + x) * Ņ32 X 8 conjugate.

24 X 14

16

= 21, the

CASE

IV.

To find the Diameter. To or from the root of the sum of the squares of the ordinate and semi-conjugate, add or fubtract the semi-conjugate, according as the less or greater absciss is used. Then, as the square of the ordinate, is to the product of the absciss and conjugate ; so is the sum or difference, above found, to the di

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The less absciss being 8, and its ordinate 14, required the diameter, fuppoing the conjugate to

be 21.

Here

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Ni X 212 + 142 + 10 6X X

= 21 X 8 X уу

14 X 14 W 212 + 282 + 21 = 3 X

= 3 X (V 3* + 42 + 3) =

7 3 x (5 + 3) = 24, the diameter required.

ÞRO

PROBLEM III.

Having given any Two Abscisses x, x, and their Ordinates y, y; to find the Diameter d to

which they belong. Multiply each absciss by the square of the ordinate belonging to the other; multiply also the square of each absciss by the square of the other's ordinate; then divide the difference of the latter products by the difference of the former; and the quotient will be the diameter to which the ordinates belong. *

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If two abfcifles be 1 and 8, and their two ordinates 4 and 14; what is the diameter to which they belong?

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By the nature of the hyperbola, dx + xx : dx + ** :: YY : yy', or dxyy + xxyy = dxYY + **YY; hence d = хх уу

2. E. D. *YY – xyy

хх үү

PROBLEM IV.

Given any Three Equidistant Ordinates a, b, c, with
their Common Distance, or Common Difference d
of their Alfcilles; to find the Diameter D.
to which they belong, together with its

Conjugate c.
* From the sum of the squares of the extreme or-
dinates, take double the square of the middle one;
and call the difference A.

From the square of the difference between A and double the said square of the middle ordinate, take 4 times the square of the product of the extreme ordinates; and call the square root of the remainder B.

Then, as A is to b, fo is the given common distance, to the diameter to which the ordinates belong.

And

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If x be the absciss to the least ordinate a, by the nature of the hyperbola we shall have these three equations, X (D + *) X *

X (Dx + xx),

сс

CC

DD

DD

CC

CC

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DD

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CE

сс

D D

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X (D+*+ d) x (x + d)

X
(D x + ** + 2 dx + Dd + dd),
X (D+* + 2d) x (x + 2d) = Х

(Dx + ** + 40x + 2D + 4dd).
And by subtracting the double of the second from the sum of
the first and third, we obtain this equation,
2bb tec=
x 2 dd; and hence

aa 2bbt oc

2 dd Again, taking the first equation from the second, leaves bb-aa

bb-aa

D+d
X (2 dxt od + dd); hence x = =

2d

CC

CC

DD

DD

СС

DD

х сс

D D

2

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