APPENDIX II.

ON THE GENERAL EQUATION OF THE SECOND DEGREE.

1. To transform the origin to the focus of the curve.
First let the co-ordinates be rectangular, and let

ay2 + bxy + cx2 + dy + ex + f = $ (x, y) = 0

be the given equation; transform the origin to the point a, ß by making

x = x + α, Y

x' +α, y y' + ß;

·· a (y' + ß)2 + b (x' + a) (y' + ß) + c (x' + a)2

+ d (y' + B) + e (x' + a) + ƒ = 0, or ay2 + bx'y' + c x22 + d'y' + e' x' + $ (a, ß) 0,

12

where d' = 2a ß + ba+d; e' = bß + 2ca + e. Next, transform the equation to polar co-ordinates by putting

··· (a sin30+bsin@cos✪+ccos30) p2+(d'sinė+e'cos☺) p+Q(a,ß)

0,

(d'2 — 4 aÞ) sin30 + (2 d'e' – 4 b Þ) sin Ocos✪ + (e22 −4cÞ) cos2 0 ;

and if a, ẞ be assumed so as to satisfy the conditions

which is the equation to a conic section from the focus, having its axis-major inclined at an angle & to the axis of x,

the two signs corresponding to the equations from the two foci respectively.

Now d'e' = 4 (ac) Þ, and d'e'

c) Þ, and d'é′ = 2bQ;

d'e' - 260 = (2 aß + ba + d) (bẞ + 2ca + e)

α

- 2b (aß2 + baß + ca2 + dẞ + ea +f)

2abß2 + (b2 + 4 ac) aß + 2bca2 + (2cd + be) a

+ (2ae+bd) ẞ + de 2b (aß + baß + ca2 + ea + dẞ+f)

dß

− (b2 — 4 a c) a ẞ + (2 cd-be) a + (2ae-bd) ẞ − (2bƒ~de) = 0;

= (2aß + ba + d)2 - 4a (aß2 + baß + ca2 + dß + ea +ƒ),

or d'2 - 4a¶ = (b2 – 4ac) a2 — (4ae — 2bd) a + d2 - 4a f.

Similarly, e-4cP =

4cÞ = ac)ẞ2
(b2 – 4a c) ẞ2 – (4cd – 2be)ß + e2 − 4cf ;

and (ah) (ẞ- k) = hk - g; from (9)

Hence hk

... a = h ±

g and μ must have the same sign, and since

ра

the two values of u have different signs; and only one of them will make a and ẞ possible.

It will afterwards be proved that in the ellipse where m
is negative; and the negative sign

is negative, bμ +

ль

A2

4a E2 = (a− h)2 + (B −k)2 = ( u + ' }) (hk − g ) = −2 \/ a'2 + m

And the co-ordinates a1, ẞ of the extremities of the axis-major are found from the equations.

a'

{μ (hk-g)}. (18)

If a', a"; ß', ß" be the two values of a and ẞ respectively determined from equations (11), a' + a′′ = 2h ; B' + ß′′ = 2k; but a', f'; a", B" are the co-ordinates of the two foci; therefore h, k are the co-ordinates of the centre.

3.

To find the values of d', e', and (a, ẞ).

4. To deduce from the above expressions the co-ordinates of the vertex, and latus rectum when b2 – 4ac = 0.

2

In this case the values of h and k become infinite;

let 2cd be K, 2ae bd H, 2bf - de

G;