through a fixed point, then the locus of its pole [A] is a fixed right line. Or, again, The intersection of any two lines is the pole of the line joining their poles; and, conversely, The line joining any two points is the polar of the intersections of the polars of these points. For if we take any two points on the polar of A, the polars of these points intersect in A.

It was proved (Art. 100) that if two lines be drawn through any point, and the points joined where they meet the curve, the joining lines will intersect on the polar of that point. Let the two lines coincide, and we derive, as a particular case of this, If through a point O any line OR be drawn, the tangents at R' and R" meet on the polar of 0; a property which might also be inferred from the last paragraph. For since R'R", the polar of P, passes through 0, P must lie on the polar of O.

P

R"

And it was also proved (Ex. 3, p. 96), that if on any radius vector through the origin, OR be taken a harmonic mean between OR and OR", the locus of R is the polar of the origin; and therefore that, any line drawn through a point is cut harmonically by the point, the curve, and the polar of the point; as was also proved otherwise (Art. 91).

Lastly, we infer that if any line OR be drawn through a point 0, and

R

T

P the pole of that line be joined to O, then the lines OP, OR will form a harmonic pencil with the tangents from 0. For since OR is the polar of P, PTRT' is cut harmonically, and therefore OP, OT, OR, OT” form a harmonic pencil.

Ex. 1. If a quadrilateral ABCD be inscribed in a conic section, any of the points E, F, O is the pole of the line joining the other two.

Since EC, ED are two lines drawn through the point E, and CD, AB, one pair of lines joining the points where they meet the conic, these lines must intersect on the polar of E; so must also AD and CB; therefore the line OF is the polar of E. In like manner it can be proved that EF is the polar of O and EO the polar of F.

E

D

Ex. 2. To draw a tangent to a given conic

A

B

section from a point outside, with the help of the ruler only.

Draw any two lines through the given point E, and complete the quadrilateral as

in the figure, then the line OF will meet the conic in two points, which, being joined to E, will give the two tangents required.

Ex. 3. If a quadrilateral be circumscribed about a conic section, any diagonal is the polar of the intersection of the other two.

We shall prove this Example, as we might have proved Ex. 1, by means of the harmonic properties of a quadrilateral. It was proved (Ex. 1, p. 57) that EA, EO, EB, EF are a harmonic pencil. Hence, since EA, EB are, by hypothesis, two tangents to a conic section, and EF a line through their point of intersection, by Art. 146, EO must pass through the pole of EF; for the same reason, FO must pass through the pole of EF; this pole must, therefore, be 0.

=

147. We have proved (Art. 92) that the equation of the pair of tangents to the curve from any point x'y' is (ax*+2hx'y' +by'2+2gx'+2ƒy'+c)(ax2+ 2hxy + by2+2gx+2fy+c) = {ax'x + h (x'y + y'x) + bý'y + g(x2 + x) +ƒ (y′ + y) + c}*. The equation of the pair of tangents through the origin may be derived from this by making x=y=0; or it may be got directly by the same process as that used Ex. 4, p. 78. If a radius vector through the origin touch the curve, the two values of p must be equal, which are given by the equation

0

(a cos2 + 2h cos sin 0 + b sin20) p2 + 2 (g cos 0 +ƒ sin 0) p + c = 0. Now this equation will have equal roots if e satisfy the equation (a cos10+2h cos sin 0 + b sin30) c = (g cos 0 +ƒ sin @)". Multiplying by p' we get the equation of the two tangents, viz. (ac-9") x2 + 2 (ch — gf) xy + (bc —ƒ”) y2 = 0.

This equation again will have equal roots; that is to say, the two tangents will coincide if

or

(ac - g3) (bc-ƒ3) = (ch — fg)3,

c (abc+2fgh — aƒ“ – bg2 — ch2) = 0.

This will be satisfied if c = 0, that is if the origin be on the curve. Hence, any point on the curve may be considered as the intersection of two coincident tangents, just as any tangent may be considered as the line joining two consecutive points. The equation will have also equal roots if

Now we obtained this equation (p. 72) as the condition that the equation of the second degree should represent two right lines. To explain why we should here meet with this equation again,

it must be remarked that by a tangent we mean in general a line which meets the curve in two coincident points; if then the curve reduce to two right lines, the only line which can meet the locus in two coincident points is the line drawn to the point of intersection of these right lines, and since two tangents can always be drawn to a curve of the second degree, both tangents must in this case coincide with the line to the point of intersection.

148. If through any point O two chords be drawn, meeting the curve in the points R', R", S, S", then the ratio of the rectangles OR. OR" will be constant, whatever be the position of the point 0, OS. OS" provided that the directions of the lines OR, OS be constant. For, from the equation given to determine p in Art. 136, it appears that

OR'. OR" =

In like manner

hence

OS. OS" =

=

OR'. OR"

=

a cos'' + 2h cos e' sine+b sin3ē'

a cos''+2h cos e' sin e' +b sin'0'

OS. OS" a cos 0+2h cos 0 sin 0 + b sin*0

But this is a constant ratio; for a, h, b remain unaltered when the equation is transformed to parallel axes through any new origin (Art. 134), and 0, e' are evidently constant while the direction of the radii vectores is constant.

The theorem of this Article may be otherwise stated thus: If through two fixed points O and O' any two parallel lines OR

and O'p be drawn, then the ratio of the rectangles

be constant, whatever be the direction of these lines. For these rectangles are

c

OR.OR"
O'p'. O'p"

will

a cos20 + 2h cos @ sin @ + b sin'0' a cos'0+2h cos✪ sin @ +b sin30 (c' being the new absolute term when the equation is transferred

to O as origin); the ratio of these rectangles, and is, therefore, independent of 0.

This theorem is the generalization of Euclid III. 35, 36.

149. The theorem of the last Article includes under it several particular cases, which it is useful to notice separately.

I. Let O' be the centre of the curve, then O'p' = O'p" and the quantity O'p'. O'p" becomes the square of the semi-diameter parallel to OR'. Hence, The rectangles under the segments of two chords which intersect are to each other as the squares of the diameters parallel to those chords.

=

II. Let the line OR be a tangent, then OR' OR", and the quantity OR.OR" becomes the square of the tangent; and, since two tangents can be drawn through the point 0, we may extract the square root of the ratio found in the last paragraph, and infer that Two tangents drawn through any point are to each other as the diameters to which they are parallel.

III. Let the line 00′ be a diameter, and OR, O'p parallel to its ordinates, then OR=OR" and O'p' O'p". Let the diameter

meet the curve in the points A, B, then

OR Ορι AO.OB AO.O'B'

=

Hence, The squares of the ordinates of any diameter are proportional to the rectangles under the segments which they make on the diameter.

150. There is one case in which the theorem of Article 148 becomes no longer applicable, namely, when the line OS is parallel to one of the lines which meet the curve at infinity; the segment OS" is then infinite, and OS only meets the curve in one finite point. We propose, in the present Article, to inquire OS' whether, in this case, the ratio will be constant. OR.OR"

Let us, for simplicity, take the line OS for our axis of x, and OR for the axis of y. Since the axis of x is parallel to one of the lines which meet the curve at infinity, the coefficient a will = 0 (Art. 138, Ex. 4), and the equation of the curve will be of the form

2hxy+by2+2gx + 2fy+c=0.

Making y = 0, the intercept on the axis of x is found to be

c

OS= ; and, making x = 0, the rectangle under the inter

2g

Now, if we transform the axes to parallel axes through any point x'y' (Art. 134), b will remain unaltered, and the new g=hy+g. Hence the new ratio will be

b

2 (hy' +g)

Now, if the curve be a parabola, h=0, and this ratio is constant; hence, If a line parallel to a given one meet any diameter (Art. 142) of a parabola, the rectangle under its segments is in a constant ratio to the intercept on the diameter.

If the curve be a hyperbola, the ratio will only be constant while y' is constant; hence, The intercepts made by two parallel chords of a hyperbola, on a given line meeting the curve at infinity, are proportional to the rectangles under the segments of the chords.

*151. To find the condition that the line λx+μy+v may touch the conic represented by the general equation. Solving for y from λx+μy+v=0, and substituting in the equation of the conic, the abscissæ of the intersections of the line and curve are determined by the equation

(aμ3 – 2hλμ + bλ3) x2 + 2 (gμ3 − hμv – fμλ + bλv) x

The line will touch when the quadratic has equal roots, or when

(aμ3 — 2hλμ + bλ2) (cμ3 — 2ƒμv + bv2) = (gμ3 – hμv – fμλ + bλv)2. Multiplying out, the equation proves to be divisible by μ3, and becomes

(bc-ƒ3) λ2 + (ca − g3) μ2 + (ab — h3) v2 + 2 (gh — aƒ) μv

+ 2 (hf − bg) vλ + 2 (ƒg − ch) λμ = 0. We shall afterwards give other methods of obtaining this equation, which may be called the tangential equation of the We shall often use abbreviations for the coefficients, and write the equation in the form

curve.

Αλ + Βμ* + * + 2 Fμν + 2 Gνλ + 21λμ = 0.

The values of the coefficients will be more easily remembered by