Refer the curve to the tangents as axes; its equation will be of the form (Art. 293)

Suppose a straight line drawn through the origin, and let its equation be (Art. 27)

Thus the distances from the origin of the points of intersection of (1) and (2) will be the values of r found from the

If r' and r" be the roots of the equation, we have

(3).

Hence for the distance (r) of the point of intersection of (2) and (5) from the origin, we have the equation

thus

is an harmonic mean between r' and r".

(6).

Since LMNO is divided harmonically, if from any point in AB we draw lines to L, N, and O, these lines with AB form

M

an harmonic pencil. A particular case is that in which the point in AB is the intersection of the tangents at N and L, which we know will meet on AB. (See Arts. 103, 186.)

336. Let A, B, C, D be four points on a conic section, and P any fifth point. Let a denote the perpendicular from P on AB, B the perpendicular from the same point on BC, y on CD, 8 on DA. Then by Art. 317 we know that wherever P may be, ay bears a constant ratio to Bd. Now AB. a twice the area of the triangle PAB

=

Similar values may be found for B, y, d. Thus

PA.PB.PC.PD

AB. CD

bears a constant ratio to

PA.PB.PC.PD

BC.AD

sin APB. sin CPD

sin BPC. sin DPA

sin APB. sin CPD

sin BPC. sin DPA;

is constant, that is, the pencil drawn

from any point P to the four points A, B, C, D, has a constant anharmonic ratio.

ANSWERS TO THE EXAMPLES.

CHAPTER I.

8. THE co-ordinates of D are (x,+x ̧) and (y,+y2). The co-ordinates of G are (x,+x+x) and (y,+y,+y).

1

2

10. Let r and be the polar co-ordinates of C. Then the angle AOC = the angle BOC; that is, 0—0,=0 ̧−0 ; .. 0=1(0,+0 ̧). Again, from the known expression for the area of a triangle (see Trigonometry, Chapter XVI.), we have

Thus r1r, sin (0,- 0 ̧) = r ̧r sin (0 — 0 ̧) + r ̧r sin (0 ̧· - 0)

2

17. The lines y=x and y = 3x.

(1) Two

(3) The 19. 4y=5x,

15. (1) The origin. (2) Two straight lines, y = x and y =-x. (3) Two straight lines, x=0 and x+y=0. (4) The axes. (5) Impossible. (6) Two straight lines, x = 0 and y = a. 16. straight lines, xa and y = b. (2) The point (a, b). point (0, a). and 3y+2x-20=0. 20. Let a be the length of the side of the hexagon; the equations are to AB, y=0; AC, y√3 = x; AD, y=x√3; AE, x = 0; AF, y + x √3 = 0; BC, y = √3 (x − a) ; BD, x = a; BE, y + √3 (x − a) = 0 ; BF, y √3 + x − a = 0; CD, y + x √3 = 2a√3; CE, y √3 + x = 3a; CF, 2y = a √3; DE, y=√3a; DF, y√3-x=2a; EF, y-x√3 = a√3. (x,, Y ̧), (x ̧, Y2), (x, y) be the angular points, the co-ordinates of the point midway between the first and second are similarly the co-ordinates of the point midway between the second and third points are known; and then the required equation can

+ x

2

2

21. If

+

1

2

х

;

1,

m2 + 1

be found by Art. 35.

22.

tan w.

24.

m2 - 1

y + = α

points whose abscisse are a +√(a2 + b2) and a

35. 90o. 36. F(0)=0 gives a system

of lines through the origin; sin 300 gives the three lines y = 0, y=x√3, y=x/3. 40. angles included by the first pair. take 4 for the origin and lines through A parallel to the two given lines as axes; let x, y, be the co-ordinates of Â, and x, y, those of C. Then it may be shewn that the equations to the three diagonals mentioned are

The second pair of lines bisect the 44. Let ABC be the triangle ;

from these equations it may be shewn that the three diagonals 45. Take O as origin and use polar equations

meet in a point.

to the given fixed straight lines. 46. Let x, be the abscissa of the point of intersection of the two lines; then the area of the triangle is (c2-c1)x,. 47. This may be solved by Art. 11. Or we may use the result of the preceding question; for by drawing a figure we shall obtain three triangles to which the preceding question applies, and the required area is the difference between two of these triangles and the third. The result is

F

3

2

2 (m ̧-m ̧) (m, — m ̧) (m ̧ — m ̧)

That sign should be taken which gives a positive result.

7. Since the required line is parallel to the line considered in Example 5, we may assume for its equation

a cos A -ẞ cos B + k = 0,

where k is some constant to be determined. Now at the middle point of AB, we have

13. (mn'-m'n) u + (nl' — n'l) v + (Im' - I'm) w = 0.

14. ab (u-v) + c (b + a) w =

15. Assume for the required equation la + mß + ny = 0; at the centre of the inscribed circle a=ẞ=y; thus l+m+ n = 0; at the centre of the circumscribed circle a, ß, y are proportional respec