we see that the coordinates of the point where the conic is met again by the line joining a"B"y" to a point on the conic a'B'y', are S'a-2Pa", S'B' -2PB", "y-2Py". These coordinates. reduce to a'B'y if the condition P=0 be fulfilled. Writing this at full length, we see that if a"B"y" satisfy the equation aaa'+bB' + cyy'+ƒ' (By' + B'y) +g (ya' + y'a) + h (aß′ + a′ß) = 0, then the line joining a"B"y" to a'B'y meets the curve in two points coincident with a'B'y'; in other words, a"ß"y" lies on the tangent at a'B'y. The equation just written is therefore the equation of the tangent.

291. Arguing, as at Art. 89, from the symmetry between, aßy, a'B'y of the equation just found, we infer that when a'B'Y is not supposed to be on the curve, the equation represents the polar of that point. The same conclusion may be drawn from observing, as at Art. 91, that P=0 expresses the condition that the line joining a’'B'y', a′′B"y" shall be cut harmonically by the The equation of the polar may be written

curve.

ả (authB+g) + B (ha+b+fy) tỷ (ga +f+c)=0. But the quantities which multiply a′, B′, y′ respectively, are half the differential coefficients of the equation of the conic with respect to a, B, y. We shall for shortness write S,, S, S, instead ds ds dS

of

da' d' dy; and we see that the equation of the polar is

In particular, if B', y both vanish, the polar of the point By is S1, or the equation of the polar of the intersection of two of the lines of reference is the differential coefficient of the equation of the conic considered as a function of the third. The equation of the polar being unaltered by interchanging aßy, a'B'y', may also be written aS,+BS +YS = 0.

292. When a conic breaks up into two right lines, the polar of any point whatever passes through the intersection of the right lines. Geometrically, it is evident that the locus of harmonic means of radii drawn through the point is the fourth harmonic to the pair of lines and the line joining their intersection to the given point. And we might also infer, from the

MM.

formula of the last article, that the polar of any point with respect to the pair of lines aß is B'a + a'ß, the harmonic conjugate with respect to a, ẞ of B'a - a'ß, the line joining aß to the given point. If then the general equation represent a pair of lines, the polars of the three points By, ya, aß,

aa+hẞ+gy = 0, ha+bB+fy = 0, ga+ƒB+cy = 0,

are three lines meeting in a point. Expressing, as in Art. 38, the condition that this should be the case, by eliminating a, B, Y between these equations, we get the condition, already found by other methods, that the equation should represent right lines, which we now see may be written in the form of a determinant, ah, g h, b, f

or, expanded,

g, f, c = 0;

abc+2fgh-af-bg" - ch2 = 0.

—

The left-hand side of this equation is called the discriminant* of the equation of the conic. We shall denote it in what follows by the letter A.

293. To find the coordinates of the pole of any line λα + μβ + νγ. Let a'B'y be the sought coordinates, then we

must have

aa' + hẞ′ + gy' =λ, ha′ +bB′ +ƒy' = μ, ga′ +ƒ3′ + cy' = v. Solving these equations for a', B', y', we get

or,

▲B' =λ(fg—ch) + μ (ca − g2) + v (gh− af),

▲y' = λ (hf − bg) + μ (gh− af) +v (ab − h2) ;

if we use A, B, C,† &c. in the same sense as in Art. 151, we find the coordinates of the pole respectively proportional to

Aλ+ Hμ+ Gv, Hλ+Bμ+ Fv, Gλ+ Fμ+ Cv. Since the pole of any tangent to a conic is a point on that tangent, we can get the condition that λa + B+ vy may touch the conic, by expressing the condition that the coordinates just found satisfy λa+μB+vy = 0. We find thus, as in Art. 285, Aλ2 + Bμ2 + Cv2 + 2Fμv + 2 Gvλ + 2Hλμ = 0.

* See Lessons on Modern Higher Algebra, Lesson XI.

† A, B, C, &c. are the minors of the determinant of the last article.

27

If we write this equation = 0, it will be observed that the coordinates of the pole are 2,, 2, 2, that is to say, the differential coefficients of Σ with respect to λ, μ, v. Just, then, as the equation of the polar of any point is aS,+BS'+yS′ = 0, so the condition that λa+μß+vy may pass through the pole of X'a + μ'ß+vy (or, in other words, the tangential equation of this pole) is λ,' + μEg' + v&g′ = 0. And again, the condition that two lines λα + μβ + νγ, λ'α + β + ν'y may be conjugate with respect to the conic, that is to say, may be such that the pole of either lies on the other, may obviously be written in either of the equivalent forms

'Σ + μ'Σ + ν'Σ = 0,

From the manner in which

λΣ' + μΣ, + vΣ, = 0.

was here formed, it appears that

Σ is the result of eliminating a', B', y', p between the equations

aa' +hẞ' + gy' + pλ = 0, ha′ + bß' +ƒý + pμ=0,

ga' +ƒß' + cy' + pv = 0, λa' + μß'+ vy = 0 ;

in other words, that Σ may be written as the determinant

λ, μ, ν, 0

=Ax2 + Bμ* + Cr2+2Fpv + 2 Gvλ + 2Hλμ.

a, h, g, λ

h, b, f, μ

g, f, c, v

Ex. 1. To find the coordinates of the pole of λa + μß +vy with respect to J(la) + √(mß) + √(ny). The tangential equation in this case (Art. 130) being

the coordinates of the pole are

ίμν + ηνλ + ηλμ = 0,

a' = mv + nμ, ß' = nλ + lv, y' = lμ + mλ.

Ex. 2. To find the locus of the pole of a +μß+vy with respect to a conic being given three tangents, and one other condition.*

Solving the preceding equations for l, m, n, we find l, m, n proportional to

λ (μβ ́ + νγ' - λα'), μ (υγ' + λα' – μβ'), ν (λα' + μβ' - νγ).

Now (la) + (mß) + √(ny) denotes a conic touching the three lines a, ẞ, y; and any fourth condition establishes a relation between l, m, n, in which, if we substitute the values just found, we shall have the locus of the pole of λa + μß + vy. If we write for λ, μ, v the sides of the triangle of reference a, b, c, we shall have the locus of the pole of the line at infinity aa + bß + cy, that is, the locus of centre. Thus the condition that the conic should touch Aa + Bß + Cy being A+B+C =0

m n

The method here used is taken from Hearn's Researches on Conic Sections.

(Art. 130), we infer that the locus of the pole of λa + μß + vy with respect to a conic touching the four lines a, ß, y, Aa + Bß + Cy, is the right line

Or, again, since the condition that the conic should pass through a'ß'y' is J(la') + √(mß') + √(ny') = 0, the locus of the pole of λa + μß + vy with respect to a conic which touches the three lines a, ẞ, y, and passes through a point a'ß'y', is

-

↓{λa' (uß + vy − λa)} + √{μß' (vy + λa − μß)} + √{vy' (\a + μß − vy)} = 0, which denotes a conic touching μβ + γ – λα, νγ + λα - μβ, λα + μβ - νγ. In the case where the locus of centre is sought, these three lines are the lines joining the middle points of the sides of the triangle formed by a, ß, y.

Ex. 3. To find the coordinates of the pole of λa+uß+vy with respect to Iẞy + mya + naß. The tangential equation in this case being, Art. 127,

12X2 + m2μ2 + n2v2 — 2mnμv — 2nlvλ – 2lmλμ = 0,

the coordinates of the pole are

a = 1 (0) - mu – ni), ở whence my + nß' = − 2lmnλ, na' + ly' = — 2lmnμ, lẞ' + ma′ = − and, as in the last example, we find l, m, n respectively proportional to

= m(mu - nv - IX), y' = n (n − 2 - mu),

2lmnv;

α' (μβ ́ + νγ' - λα'), β' (υγ' + λα - μβ'), γ' (λα' + μβ' - νγ'). Thus, then, since the condition that a conic circumscribing aßy should pass through a fourth point a'ẞ'y' is +. + = 0, the locus of the pole of λa + μß + vy, with regard to a conic passing through the four points, is

a

m

n

which, when the locus of centre is sought, denotes a conic passing through the middle points of the sides of the triangle. The condition that the conic should touch Aa + Bẞ+ Cy being √(Al) + √(Bm) + √(Cn) = 0, the locus of the pole of λa + μß +vy, with regard to a conic passing through three points and touching a fixed line, is

Δα (μβ + γ - λα)} + {{Ββ (vy + λα - μβ)} + \ Cy (λα + μβ - νγ) = 0, which, in general, represents a curve of the fourth degree.

294. If a"B"y" be any point on any of the tangents drawn to a curve from a fixed point a'B'y', the line joining a′ß'y', a′′B′′y”. meets the curve in two coincident points, and the equation in 7: m (Art. 290), which determines the points where the joining line meets the curve, will have equal roots.

To find, then, the equation of all the tangents which can be drawn through a'B'y', we must substitute la+ma', lẞ+mß', ly+my in the equation of the curve, and form the condition that the resulting equation in 7:m shall have equal roots.

Thus (see Art. 92) the equation of the pair of tangents to a conic is SS P2, where

S=aa2 + &c., S′ = aa22 + &c., P= aaa' +&c.

This equation may also be written in another form; for since any point on either tangent through a'B'y evidently possesses the property that the line joining it to a'B'y touches the curve, we have only to express the condition that the line joining two points (Art. 65)

a (B'y” — B′′y') + ß (ya” — y′′a) + y (a ́B” — a′′B') = 0 should touch the curve, and then consider a"B"y" variable, when we shall have the equation of the pair of tangents. In other words, we are to substitute By - B'r, vá - y'a, aß′-aß for λ, μ, in the condition of Art. 285,

Ax2 + Bμ2 + Cv2 + 2Fμv+2 Gvλ + 2Hλμ = 0. Attending to the values given (Art. 285) for A, B, &c., it may easily be verified that

2

(aa2 + &c.) (aa2 + &c.) - (aaa' + &c.)2 = A (By' — B'y)2 + &c.

Ex. To find the locus of intersection of tangents which cut at right angles to a conic given by the general equation (see Ex. 4, p. 169).

We see now that the equation of the pair of tangents through any point (Art. 147) may also be written

A (y - y')2 + B (x − x′)2 + C′ (xy' — yx')2

— 2F (x − x′) (xy' — yx′) + 2G (y − y′) (xy' — x'y) — 2H (x − x′) (y — y') = 0. This will represent two right lines at right angles when the sum of the coefficients of x2 and y2 vanishes, which gives for the equation of the locus

C (x2+ y2) - 2Gx2Fy + A + B = 0.

This circle has been called the director circle of the conic. When the curve is a parabola, C = 0, and we see that the equation of the directrix is Gx + Fy = } (A + B).

295. It follows, as a particular case of the last, that the pairs of tangents from By, ya, aß are

By2 + CB2 – 2Fßy, Ca2 + Ay2 - 2 Gya, Aß2 + Ba2 – 2Haß, as indeed might be seen directly by throwing the equation of the curve into the form

(aa+hẞ+gy)2 + (CB2 + By2 − 2Fßy) = 0.

Now if the pair of tangents through By be ẞ-ky, B-k'y, it

B appears from these expressions that kk' = C'

and that the corre