1269. Each diameter of a given parabola meets a fixed straight line, and from their common point is drawn a straight line making a given angle with the tangent corresponding to the diameter: the envelope is a parabola, degenerating when the given angle is equal to the angle which the fixed straight line makes with the axis.

1270. From the point where a diameter of a given conic meets a fixed straight line is drawn a straight line inclined at a given angle to the conjugate diameter: the envelope is a parabola. If the constant angle be made with the first diameter, the envelope is another parabola. [The equations of the conic and straight line being

and t the tangent of the given angle, the envelopes are respectively {pa (y − tx) + qb (x + ty) + t (a2 − b2) }2 = 4ab {p (x + ty) — a} {q (y — tx) — b}, and {bp (x+ty) — aq (y — tx)}2 + 4ab {q (x + ty) — bt} {p (y − tx) + at} = 0.]

1271. A fixed point A is taken on a given circle, and a chord of the circle PQ is such that PQ-e (APAQ): prove that the envelope of PQ is a circle of radius a (1-e) touching the given circle at A, a negative value of the radius meaning external contact.

1272. A fixed point A is taken within a given circle, and a chord of the circle PQ is such that PQ-e (AP+AQ): prove that the envelope of PQ is a circle coaxial with the given circle and the point A and whose radius is √(a2 – e°c2)(1 − e2), where a is the radius of the given circle and c the distance of A from its centre.

1273. One given circle U lies within another V, and PQ is a chord of V touching U, S the interior point circle coaxial with U and V, PSP a chord of V prove that the envelope of QP' is a third coaxial circle such that the tangents drawn from any point of it to U, V are in the ratio a2 – c2 : a2, where a is the radius of U and c the distance between their centres.

1274. A circle passes through two fixed points A, B, and a tangent is drawn to it at the second point where it meets a fixed straight line through A: the envelope of this tangent is a parabola, whose focus is B, whose directrix passes through A, and whose axis makes with BA an angle double that which the fixed straight line makes with BA. [If the two points be (±a, 0), and y cos a = straight line, the envelope is

--

= (x − a) sin

(x + a)2 + y2 = {(x − a) cos 2a + y sin 2a}.]

α

the given

1275. Through any point 0 on a fixed tangent to a given parabola is drawn a straight line OPTP meeting the parabola in P, P' and a given straight line in 7, and OT is a mean proportional between OP, OP: prove that PP either passes through a fixed point or envelopes a parabola.

1276. The envelope of the polar of the origin with respect to any circle circumscribing a maximum triangle inscribed in the ellipse a2y" + b2x2 = a'b' is

[If be the fourth point in which the circumscribing circle meets the ellipse, the equation of the polar will be

1277. A chord of a given conic is drawn through a given point, another chord is drawn conjugate to the former and equally inclined to a given direction: prove that the envelope of this latter chord is a parabola.

1278. A triangle is self-conjugate to the circle (x - c) + y = b2, and two of its sides touch the circle x2 + y2=a: prove that if the equation of the third side be p (x−c) +qy = 1, p, q will be connected by the equation

p2 (a2 + b3) + q3 (a2 + b2 − c2) + 2pc + c2 − 2a2 = 0 ;

and find the Cartesian equation of the envelope.

1279. A triangle ABC is inscribed in the circle x2+ y2 = a3, and A is the pole of BC with respect to the circle (x-c) + y = b2: prove that AB, AC envelope the conic

1280. From a fixed point 0 are drawn tangents OP, OP' to one of a series of conics whose foci are given points S, S': prove that (1) the envelope of the normals at P, P is the same as the envelope of PP', (2) the circle OPP' will pass through another fixed point, (3) the conic OPP'SS" will pass through another fixed point.

1281. A chord of a parabola is drawn through a fixed point and on it as diameter a circle is described: prove that the envelope of the polar of the vertex with respect to this circle is a conic which degenerates when the fixed point is on the tangent at the vertex.

a

[This conic will be a circle for the point (-2, 0), and a rectangular

hyperbola when the point lies on the parabola y2 = 4a (2x − a).]

==

1282. The centre of a given circle is C and a diameter is AB, chords AP, PB are drawn and perpendiculars let fall on these chords from a fixed point 0: prove that the envelope of the straight line joining the feet of these perpendiculars is a conic whose directrices are AB and a parallel through O, whose excentricity is CP: CO, and whose focus corresponding to the directrix through O lies on CO.

1283. From the centres A, B of two given circles are drawn radii AP, BQ whose directions include a constant angle 2a: prove that the envelope of PQ is a conic whose excentricity is

C

Ja2+b2 - 2ab cos 2a

where a, b are the radii and c the distance AB.

[The conic is always an ellipse when one circle lies within the other, and always an hyperbola when each lies entirely without the other; when the circles intersect, the conic is an ellipse if 2a be greater than the angle subtended by AB at a common point, reduces to the two common points when 2a has that critical value, and is an hyperbola for any smaller numerical value. When 2a = 0 or π, the envelope degenerates to a point. For different values of a, the foci of the envelope lie on a fixed abc and whose centre divides AB externally in the a2 - 12'

circle of radius ratio a: b.]

1284. Two conjugate chords AB, CD of a conic are taken, P is any point on the conic, PA, PB meet CD in a, b, O is another fixed point, and Oa, Ob meet PB, PA in Q, R: prove that QR envelopes a conic which degenerates if ( lie on AB or on the conic.

1285. The point circles coaxial with two given equal circles are S, S', a straight line parallel to SS' meets the circles in H, H' so that SH = S'H', and with foci II, II' is described a conic passing through S, S': prove that its directrices are fixed and that its envelope is a conic having S, S' for foci.

1286. Two conics U, V osculate in O and PP' is the remaining common tangent, PQ, P'Q' are drawn tangents to V, U respectively : prove that PP', QQ' and the tangent at O meet in a point, and that, if from any point on PP' be drawn other tangents to U, V, the straight line joining their points of contact envelopes a conic touching both curves at Ø and touching U, V again in Q, Q respectively.

1287. Normals PQ, PQ' are drawn to the parabola y = 4ax from a point P on the curve: prove that the envelope of the circle PQQ' is the

which is the pedal of the parabola ya (x-2a) with respect to the origin A.

[The chord QQ' always passes through a fixed point C(− 2a, 0), and if S be the focus of the given parabola, C, S are single foci and A a double focus of the envelope: for a point P on the loop, C'P = AP + 2SP, and for a point on the sinuous branch, CP = 2SP – AP, so that AP may be regarded as changing sign in vanishing.]

VIII. Area! Co-ordinates.

[In this system the position of a point P with respect to three fixed points A, B, C not in one straight line is determined by the values of the ratios of the three triangles PBC, PCA, PAB to the triangle ABC, any one of them PBC being esteemed positive or negative according as P and A are on the same or on opposite sides of BC. These ratios being denoted by x, y, z will always satisfy the equation x + y + z =1. A point is completely determined by the ratios of its areal co-ordinates (Y: Y: Z) or by two equations, as lx = my = nz. It is sometimes convenient to use trilinear co-ordinates, x, y, z being then the distances of the point from the sides of the triangle of reference ABC and connected by the equation ax + by + cz = 2K, where a, b, c are the sides and K the area of the triangle ABC. A point would obviously be equally well determined by x, y, z being any fixed multiples of its areal or trilinear co-ordinates, a relation of the form Ax+ By + Cz = 1 always existing. In the questions under this head areal co-ordinates will generally be taken for granted.

The general equation of a straight line is px + qy+rz = 0, and P, q, rare proportional to the perpendiculars from A, B, C on the straight line, sign of course being always regarded. When p, q, r are the actual perpendiculars, px+qy+rz is the perpendicular distance from the line of the point whose areal co-ordinates are x, Y, Z.

The condition that the straight lines p,x+q,y+r,z=0, p ̧x+q ̧¥+r ̧z=0 shall be parallel is

If p1 = q=r1, or if p, q, r,, both these equations are true. The straight line x + y + z = 0, the line at infinity, may then be regarded as both parallel and perpendicular to every finite straight line, the fact being that the direction of the line at infinity is really indeterminate.

In questions relating to four points it is convenient to take the points to be (X, ± Y, ± Z), or given by the equations lxmy=nz; and similarly to take the equations of four given straight lines to be px±qy±rz= = 0.

The general equation of a conic is

u = ax2 + by + cz2 + 2fyz + 2grx + 2hxy = 0;

The special forms of this equation most useful are

(1) circumscribing the triangle of reference (a, b, c = 0)
fyz+gzx+hxy = 0;

(2) inscribed in the triangle of reference

(lx) + (my)1 + (nz)1 = 0;

(3) touching the sides AB, AC at the points B, C

kx2=yz;

but when this form is used it is often better to take such a multiple of the ratio APBC : AABC for x as to reduce the equation to the form

x2 = yz;

(4) to which the triangle is self-conjugate (f, g, h=0)

lx2 + my3 + nz3 = 0;

here again it is often convenient to use such multiples of the triangle ratios as to give us the equation

x2 + y2+~2 = 0.

As a general rule, when metrical results are wanted, it will be found simpler to keep to the true areal or trilinear co-ordinates.

The form (4) is probably the most generally useful. We may denote any point on such a conic by a single variable, as with the excentric angle in the case of a conic referred to its axes, which is indeed a particular case of this form. Thus any point on the conic x2+ y2+ z = 0 may be represented by the equations