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Let AFBDCE be the conic; take ABC as the triangle of reference, and let the equation of the conic be


Let the equation of AE be ß=n,y of AF be y=mxß,

g= lyd, of BD

= nir, CD m,ß, of CE ... B=1,a. Then, since D lies in the conic (1), we have +um, + vn, =0,

..... ale + + vn, = 0, F

ul+ um, +v=0, 1, m, n, whence lo, 1, n, = 0...

(2) l, m, 1, is the necessary condition that the six points A, F, B, D, C, E may lie in a conic.

Again, if the pairs of opposite sides intersect in points lying in a straight line, let the equation of that straight line be pa + +rry= 0. Then, since BF and CE intersect in this line, we have P +91+vle = 0, CD and AF

pm, + 9 + vm, = 0, AE and BD

pn, + gn, + v= 0, 1, la, la whence


........ (3)

AF .......


1, mg


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is the condition that these points of intersection may lie in the same straight line. But (2) and (3) are identical. Hence the proposition is proved.

From Pascal's Theorem many interesting consequences may be deduced. Thus, if the point F coincide with

A, D with B, E with C, then AF, BD, CE become the tangents at A, B, C respectively, and we obtain the theorem enunciated in Ex. 1. Chap. II. Again, by supposing D to coincide with B, and E with C, we readily obtain the following theorem: “If the opposite sides of a quadrilateral, inscribed in a conic, be produced to meet, and likewise the pairs of tangents at opposite angles of the quadrilateral, the four points of intersection will lie in the same straight line."

And, by supposing F to coincide with A, we obtain a geometrical construction, by which, having given five points of a conic, we can draw a tangent at any one of them. For, since AF then becomes the tangent at Å, we see that, if AE, DB be produced to meet in G, AB, EC in H, and GH intersect CD in I, then AI will be the tangent at A.


1. Prove that

, b, a, d, c c, d, a, b 6 d, c, b, a

=(a+b+c+d) (a 6+c-d)(a-6-c+d)(a+b-c-d).

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1. We may now proceed to the discussion of the general equation of the second degree, which we shall express under the form,

ua? + vß2 + 2y + 2u'ßry + 20'ya + 2w'aß=0.

This we may write, for shortness, $ (a, b, y) = 0.

This equation, as we have shewn (Art. 1, Chap. III), represents a conic section.

2. To find the point in which a straight line, drawn in a given direction through a given point of the conic, meets the conic again.

Let f, g, h, be the co-ordinates of the given point, a, b, y those of any other point whatever. Then, for all points of the straight line joining these two, the quantities

a-f, B-9, y-h,

will bear constant ratios to one another. Let these ratios be denoted by p:q:r, so that we have a-f-B-9_n-h

=S, suppose. P 2


To find where the line again meets the conic, we must substitute in the equation of the conic

f+ps for a, g+qs for B, h +rs for y. We thus get, arranging the result according to ascending powers of 8, $(f, g, h) + 2 {(up + w'q+v'r) f + (w'p+vq+u'r) g

+ (u'p+u'q + wr) h} s + (p, q, r) se = 0. The two roots of this equation, considered as a quadratic in s, determine the two points where the line meets the conic.

Now, since (f, g, h) is, by supposition, a point on the conic, it follows that $ (f, g, h) must be itself=0. Hence, one of the two values of s, given by the above equation, will = 0, as ought to be the case, this value corresponding to the point f, g, h itself. The value of s, corresponding to the other point of intersection, will then be

(up + w'q + v'r).f+(w'p + vq+u'r)g+up+u'q+wr)h - 2

(p, q, r)
Hence, the values of a, B, Y, may be determined.

To this value of s, we shall hereafter have occasion to refer.

3. To find the equation of the tangent at a given point.

If the two points in which a straight line meets the conic be indefinitely close together, the value of s, investigated in Art. 2, must be = 0. This gives (up + w'q + vr) f + (w'p +vq + u'r) g + (up+uq + wr) h= 0, or, (uf+w'g + v'h) p+ (w'f + vg + u'h) 2 + (0f + u'g + wh) r=0. Hence, since, for every point on the line required,




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