normal at O, and that the straight lines joining Q', P' to the centre of the circle OPQ will cut off one-third from OP, OQ respectively.

[Q', P will be the centres of perpendiculars of the vanishing triangles 0, 0, P; 0, 0, Q respectively.]

875. The length of a chord of a R. H. which is normal at one extremity is equal to the corresponding diameter of curvature.

[Take on the hyperbola three contiguous points ultimately coincident and consider the centre of the circumscribed circle, centroid, and centre of perpendiculars of the infinitesimal triangle.]

876. A diameter PP being taken, a circle is drawn through P touching the hyperbola in P; prove that this circle is equal to the circle of curvature at P, and that if PI be the diameter of curvature at P, PR the common chord of the hyperbola and circle of curvature, RI will be equal and parallel to PP'.

877. A triangle is inscribed in a circle, and two parabolas drawn touching the sides with their foci at ends of a diameter of the circle: prove that their axes are asymptotes of a rectangular hyperbola passing through the centres of the four circles which touch the sides.

878. Three tangents are drawn to a R. H. such that the centre of the circle circumscribing the triangle lies on the hyperbola: prove that the centre of the hyperbola will lie on the circle; and that at any common point tangents drawn to the two curves pass through the points of contact of a common tangent.

879. A circle meets a R. H. in points P, P, Q, Q' and P, P′ are ends of a diameter of the hyperbola: prove that the tangents to the hyperbola at P, P' and to the circle at Q, Q' are parallel, and the tangents to the circle at P, P' and to the hyperbola at Q, Q' all meet in one point.

880. The tangent at a point of a R. H. and the diameter perpendicular to this tangent being drawn; prove that the segments of any other tangent from its point of contact to these two straight lines subtend supplementary angles at the point of contact of the fixed tangent.

881. The normal at a point P meets the curve again in Q; RR' is a chord parallel to this normal: prove that the points of intersection of QR, PR' and of QR', PR lie on the diameter at right angles to CP.

с

882. A triangle ABC is inscribed in a R. H. and its sides meet one asymptote in a, b, c and the other in a', b', c' respectively: through a, b, c are drawn straight lines at right angles to the corresponding sides of the triangle: prove that these meet in a point 0, and, O' being similarly found from a', b', c', that OO' is a diameter of the circle ABC.

CONIC SECTIONS, ANALYTICAL.

CARTESIAN CO-ORDINATES.

I. Straight Line, Linear Transformation, Circle.

[In any question relating to the intersections of a curve and two straight lines, it is generally convenient to use one equation representing both straight lines. Thus, to prove the theorem: “ Any chord of a given conic subtending a right angle at a given point of the conic passes through a fixed point in the normal at the given point;" we may take the equation of the conic referred to the tangent and normal at the given point

ax2+2hxy+by2 = 2x;

the equation of any pair of straight lines through this point at right angles to each other is

x2 + 2λxy — y2 = 0 ;

and at the points of intersection

(a + b) x2 + 2 (h + λb) xy = 2x;

or, at the points other than the origin,

(a + b) x + 2 (h + λb) y = 2,

which is therefore the equation of a chord subtending a right angle at the origin. This passes through the point y = 0, (a+b) x = 2; a fixed point on the normal.

If two points be given as the intersections of a given straight line and a given conic the equation of the straight lines joining these points to the origin may be formed immediately, since it must be a homogeneous equation of the second degree in x, y. Thus the straight lines joining the origin to the points determined by the equations

ax2 + 2hxy + by2 = 2x,

px + qy = 1,

are represented by the equation

ax2 + 2hxy + by2 = 2x (px + qy),

and will be at right angles if a+b=2p, or if the straight line a somewhat different

px+qy=1 pass through the point (2,0),

mode of proving the theorem already dealt with. In general the equation of the straight lines joining the origin to the two points determined by the equations

2

is ax2 + 2hxy + by2 — — (px + qy) (gx +fy) + ~ (px + qy)" = 0.

The results of linear transformation may generally be obtained from the consideration that, if the origin be unaltered, the expression

must be transformed into

x2 + 2xy cos w + y2

X+2XY cos 0 + Y2,

if (x, y), (X, Y) represent the same point and w, be the angles between the co-ordinate axes in the two systems respectively. Thus if u = ax2+by+c+2fy + 2gx+2hxy

be transformed into

U = AX2 + BY2 + c + 2FY + 2GX + 2HXY,

then λ (x2 + y2+ 2xy cos w) + u must be transformed into

λ(X2+ Y2 + 2XY cos n) + U,

and if A have such a value that the former be the product of two linear factors, so also must the latter; hence the two quadratic equations in A

and

c (λ + a) (λ + b) + 2 fg (λ cos w + h)

= (λ + a) ƒ2 + (λ + b) g2 + c (λ cos w + h)3,

c (λ + A) (λ + B) + 2FG (λ cos + H)

= (X + A) F2 + (A + B) G3 + c (λ cos + H)3

must coincide; and thus the invariants may be deduced. Also, by the same transformation,

λ (x2+ y2+ 2xy cos w) + ax2+by+2hxy

must be transformed into

λ (X2 + Y2 + 2XY cos Q) + AX2 + BY2 + 2HXY,

and if A have such a value that the former is a square, so must the latter; hence the equations

(λ + a) (λ + b) = (λ cos w + h)2,

(X+A) (A + B) = (λ cos + H)',

One special form of the equation of a circle is often useful it is

(x − x ̧) (x − x ̧) + (y − y ̧) (y − y,) = 0,

where (x, y), (x, y) are the ends of a diameter, and the axes rectangular. The corresponding equation when the axes are inclined at an angle w is obtained by adding the terms

{(x − x ̧) (y − Y2) + (x − x ̧) (y − y ̧)} cos w ;

each equation being found at once from the property that the angle in a semicircle is a right angle. In questions relating to two circles, it is generally best to take their equations as

the axis of being the radical axis, and k negative when the circles intersect in real points.]

883. The equation of the straight lines which pass through the origin and make an angle a with the straight line x + y = 0 is

x2+2ry sec 2a + y2 = 0.

=

884. The equation be- 2xy + ay 0 represents two straight lines at right angles respectively to the two whose equation is

ax2 + 2hxy + by3 = 0.

If the axes of co-ordinates be inclined at an angle w, the equation will be (a + b − 2h cos w) (x2 + y2 + 2xy cos w) = (ax2 + 2hxy + by3) sin2 w.

885. The two straight lines

2

x2 (tan + cos 0) — 2xy tan 0 + y2 sin3 0 = 0 make with the axis of x angles a, ẞ such that tan a

886. The two straight lines.

tan ẞ= 2.

(x+y) (cos' sin' a + sin' 0) = (x tan ay sin 0)

include an angle a.

887. The two straight lines.

sin' a cos2 + 4xy sin a sin 0 + y2 {1 cosa

include an angle a.

(1 + cos a)2 cos2 0} = 0

888. Form the equation of the straight lines joining the origin to the points given by the equations

(x − h)2 + (y − k)2 = c2, kx + hy = 2hk,

and prove that they will be at right angles if h+kc. Interpret geometrically.

889. The straight lines joining the points given by the equations

ax2 + by3 +c+2fy + 2gx + 2hxy = 0, px + qy = 1,

to the origin will be at right angles if

a+b+ 2 (fq+gp) + c (p2 + q3) = 0;

and the locus of the foot of the perpendicular from the origin on the line px + qy = 1 is (a+b) (x2 + y2) + 2ƒy + 2gxr+c=0: also the same is the locus of the foot of the perpendicular from the point

is the parts of two straight lines at right angles to each other which include one quadrant.

[The equation gives y=1+x when a is positive and y = 1−x when x is negative.]

891. The formulæ for effecting a transformation of co-ordinates, not necessarily rectangular, are

being the angles between the co-ordinate axes in the two cases.