6.

To find the equation of a conic which touches the three sides of the triangle of reference.

are

The co-ordinates of the sides of the triangle of reference

q=0, r=0 for BC,

r =0, p=0 for CA,

p=0, q=0 for AB.

Hence, the equation of the required conic must be satisfied whenever two out of the three co-ordinates p, q, r are = 0. It must therefore be of the form

Lqr+Mrp+Npq=0.

The equations of the points of contact are

These may be established as follows: If in the given equation we make Mr+ Nq = 0, we obtain either q = 0, or r=0. It hence appears that the tangents drawn through the point Mr+Nq=0, pass either through the point q=0, or through the point r=0. But the three points

Mr+ Nq=0, q=0, r=0,

lie in the same straight line; hence the tangents drawn from Mr+Nq=0 coincide, that is, it is the point of contact of the tangent for which q=r=0. Similarly for the other two points of contact.

It will hence appear, by reference to the equations of the points of contact of the inscribed circle, given in Art. 2, that that circle is represented by the equations

7. To find the equation of a conic circumscribed about the triangle of reference.

The equations of the angular points of the triangle of reference are p=0, q=0, r=0. Now, since each of these points lies on the curve, the two tangents drawn through any one of them must coincide, hence when any one of these quantities is put = 0, the remaining equation must have two equal roots. The required equation will therefore be of the

form

L3p2 + M3q3 +N2r3 — 2MNqr — 2NLrp — 2LMpq = 0.

The co-ordinates of the several tangents at the angular points will be given by the equations

p=0, Mq- Nr =0,

q=0, Nr - Lp = 0,

r=0, Lp-Mq = 0.

If the conic be a circle, the tangent at A will be determined by the equations

Similar equations holding for the other two tangents, the equation of the circumscribing circle will be

2

a2p2 + bʻq2+cʻr2 - 26°c qr - 2c3a3rp - 2a b3pq=0,

which may be reduced to

± ap1± bq* ± cr3 = 0.

8. By investigations similar to those in Chap. IV. Art. 8, it may be shewn that the equation of the pole of the line (f, g, h) with respect to the conic

is

$ (p, q, r) = up2 +vq2 + wr2 + 2u'qr +2v'rp + 2w'pq = 0,

(uf+w'g+v'h) p + (w'f+vg+u'h) q + (v'f+u'g+wh) r = 0. Now, the centre is the pole of the line at infinity, which is given by the equations p=q=r.

The equation of the centre is therefore

(u + v' + w') p + (u' + v + w') q + (u' + v' + w) r = 0.

If the conic be a parabola, it touches the line at infinity; the condition that it should be a parabola is therefore

u+v+w+Qu' + 2v′ + 2w' = 0.

9. The two points in which the conic is cut by the line (f, g, h) are represented by the equation

Hence, the two points in which it is cut by the line at infinity are given by

4(u+v+w+2u'+2v'+2w') (up3+vq3+wr2+2u'qr+2v'rp+2w'pq) −{(u+v'+w') p+(u'+v+w')q+(u'+v'+w)r}2=0.

Hence may be deduced the equation of the two points at infinity through which all circles pass. For these are the same for all circles. Now, for the inscribed circle they are obtained by putting

u=v=w=0, 2u's-a, 2v's-b, 2w' = s — c.
The equation then becomes

4s {(s—a) qr + (s− b) rp + (s — c) pq} − (ap+bq+cr)2= 0,

or a2p2+b2q2+c2r2-2bcqrcos A-2 carp cos B-2abpqcos C=0,

which may also be written

a2 (p − q) (p − r) + b2 (q − r) (q − p) + c2 (r − p) (r − q) = 0,

the equation of the two circular points at infinity.

10. Every circle may therefore be represented by the equation

a2 (p − q) (p − r) + b2 (q− r) (q − p) +c2 (r − p) (r − q)

· (lp + mq + nr)2 = 0,

where lp+mq+nr=0 is the equation of the centre.

Hence may also be deduced the conditions that the equation up2 + vq2 + wr2 + 2u'qr + 2v'rp+2w'pq = 0

may represent a circle. For, comparing this equation with that just obtained, we get

Putting each member of these equations=k, they may be written

a2 - ukl2, b2 - vk m2, c2-wkn2,

=

or (vw-u') k2 (vc2+wb3 + 2u'bc cos A) k + b c sin2 A=0. · b2c2

Again, we get

(a2 — uk) (bc cos A + u'k) + (ca cos B+ v'k) (ab cos C+w'k) =0, or (v'w' —uu') k2+{a(au'+ bcos C.v'+ccos B. w') - bccos A.u} k

- a2bc sin B sin C=0.

Now, since b'c' sin' A= abc sin B sin C=(2A), these equations may be written under the form

(vw — u'2) k2 — (vc2 + wb2 + 2u'bc cos A)k + 4A2 = 0, (v'w'—uu')k2+{a (au'+bv'cos C+cw'cos B)-bccos A'.u} k+4A2=0. Combining these with the four similar equations, we get

- {u (b+c2-2bc cos A) + va2+wa2 +2a3u'

+2cv' (c cos B+ b cos C) +2aw' (b cos C+ c cos B)} = 0,

Two other corresponding expressions may of course be obtained for k, and the required condition is therefore

1. A parabola is described about a triangle so that the tangent at one angular point is parallel to the opposite side; shew that the square roots of the perpendiculars on any tangent to the curve are in arithmetical progression.

2. A conic is circumscribed about a triangle such that the tangent at each angular point is parallel to the opposite side; shew that, if p, q, r be the perpendiculars from the angular points on any tangent,

3. Shew that the equation of the centre of this conic is

p+q+r=0.