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x cosp'+y sino', or = ('). Hence, the angle between two tangents to a parabola is half the angle which the points of contact subtend at the focus; and again, the locus of the intersection of tangents to a parabola, which contain a given angle, is a hyperbola with the same focus and directrix, and whose eccentricity is the secant of the given angle, or whose asymptotes contain double the given angle (Art. 167).

282. Any two conics have a common self-conjugate triangle. For (see Ex. 1, p. 143) if the conics intersect in the points A, B, C, D, the triangle formed by the points E, F, O, in which each pair of common chords intersect, is self-conjugate with regard to either conic. And if the sides of this triangle be a, ß, Y, the equations of the conics can be expressed in the form

ca +b + c = 0,
tay a'a2 + b'ß2 + c'y2 = 0.

We shall afterwards discuss the analytical problem of reducing the equations of the conics to this form. If the conics intersect in four imaginary points, the lines a, B, y are still real. For it is obvious that any equation with real coefficients which is satisfied by the co-ordinates x'+x” √(−1), y' + y′′ √(−1), will also be satisfied by x' — x' √√(−1), y' — y′′ √√(— 1), and that the line joining these points is real. Hence the four imaginary points common to two conics consist of two pairs x' ±x′′ √(− 1), y'±y′′ √(−1); x"±x'' √(-1), y"±y"" √(-1). Two of the common chords are real and four imaginary. But the equations of these imaginary chords are of the form L± M√(−1), L'+ M' √(-1), intersecting in two real points LM, L'M'. Consequently the three points E, F, O are all real.

If the conics intersect in two real and two imaginary points, two of the common chords are real, viz. those joining the two real and two imaginary points; and the other four common chords are imaginary. And since each of the imaginary chords passes through one of the two real points, it can have no other real point on it. Therefore, in this case, one of the three points E, F, O is real and the other two imaginary; and one of the sides of the self-conjugate triangle is real and the other two imaginary.

Ex. 1. Find the locus of vertex of a triangle whose base angles move along one conic, and whose sides touch another. [The following solution is Mr. Burnside's.]

Let the conic touched by the sides be 22+ y2 - z2, and the other ax + by2 — cz2, Then, as at Ex. 1, p. 94, the co-ordinates of the intersection of tangents at points α, Y, are cos(a + y), sin (a + y), cos (a − y); and the conditions of the problem give a cos2 (a + y) + b sin2 (a + y) = c cos2 (a − y) ;

or

(a + b − c) + (a − b −c) cosa cos y + (b − c − a) sin a sin y = 0; In like manner

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and since the co-ordinates of the point whose locus we seek are cos (a + ẞ), sin (a + B), cos (a - ẞ), the equation of the locus is

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Ex. 2. A triangle is inscribed in the conic x2 + y2 = z2; and two sides touch the conic ax2 + by2 = cz2; find the envelope of the third side.

Ans. (ca + ab- bc)2 x2 + (ab + bc − ca)2 y2 = (bc + ca — ab)2 z2.

ENVELOPES.

283. If the equation of a right line involve an indeterminate quantity in any degree, and if we give to that indeterminate a series of different values, the equation represents a series of different lines, all of which touch a certain curve which is called the envelope of the system of lines. We shall illustrate the general method of finding the equation of an envelope, by proving, independently of Art. 270, that the line μ3L-2μR+M, where is indeterminate, always touches the curve LM-R". The point of intersection of the lines answering to the values μ and μ+k, is determined by the two equations

μ

μ3L− 2μR+M=0, 2(μL−R)+kL=0 ;

the second equation being derived from the first by substituting μk for μ, erasing the terms which vanish in virtue of the first equation, and then dividing by k. The smaller k is, the more nearly does the second line approach to coincidence with the first; and if we make k=0, we find that the point of meeting of the first line with a consecutive line of the system is determined by the equations

or,

μ3L−2μR+M=0, μL-R=0;

what comes to the same thing, by the equations

μL-R=0, μR-M=0.

Now since any point on a curve may be considered as the intersection of two of its consecutive tangents (p. 144), the point where any line meets its envelope is the same as that where it meets a consecutive tangent to the envelope; and therefore the two equations last written, determine the point on the envelope which has the line L-2μR+M for its tangent. And by eliminating μ between the equations we get the equation of the locus of all the points on the envelope, namely LM=R".

A similar argument will prove, even if L, M, R do not represent right lines, that the curve represented by μ3L−2μR+M, always touches the curve LM-R".

The envelope of L cos+ M sino-R, where & is indeterminate, may be either investigated directly in like manner; or may be reduced to the preceding by assuming tanμ, when on substituting

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and clearing of fractions, we get an equation in which μ only enters in the second degree.

284. We might also proceed as follows: The line

μ2L-2μR+M

is obviously a tangent to a curve of the second class (see note, p. 142); for only two lines of the system can be drawn through a given point: namely, those answering to the values of μ determined by the equation

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where L', R', M' are the results of substituting the co-ordinates of the given point in L, R, M. Now these values of μ will evidently coincide, or the point will be the intersection of two consecutive tangents, if its co-ordinates satisfy the equation LM=R3. And, generally, if the indeterminate μ enter algebraically and in the nth degree, into the equation of a line, the line will touch a curve of the nth class, whose equation is found by expressing the condition that the equation in μ shall have equal roots.

Ex. 1. The vertices of a triangle move along the three fixed lines a, ß, y, and two of the sides pass through two fixed points a'ß'y', a"ẞ"y", find the envelope of the third side. Let auß be the line joining to aẞ the vertex which moves along y, then the equations of the sides through the fixed points are

y' (a + μß) − (a' + μß') y = 0, y" (a + μß) − (a” + μß") y = 0.

And the equation of the base is

(a' + μß') y'a + (a” + μß") μɣ'ß — (a′ + μß') (a” + μß′′) y = 0,

for it can be easily verified, that this passes through the intersection of the first line with α, and of the second line with ß. Arranging according to the powers of μ, we find for the envelope

(aß'y" + By'a" — ya'ß" — ya”ß')2 = 4a′ß" (ay" — a′′z) (By' — B′y).

This example may also be solved by arranging according to the powers of a, the equation in Ex. 3, p. 49.

Ex. 2. Find the envelope of a line such that the product of the perpendiculars on it from two fixed points may be constant.

Take for axes the line joining the fixed points and a perpendicular through its middle point, so that the co-ordinates of the fixed points may be y = 0, x = c; then if the variable line be y — mx + n = 0, we have by the conditions of the question

or

but

therefore

and the envelope is

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Ex. 3. Find the envelope of a line such that the sum of the squares of the perpendiculars on it from two fixed points may be constant.

Ans.

2x2

2y2 + b2 2c2 b2

Ex. 4. Find the envelope if the difference of squares of perpendiculars be given.

= 1.

Ans. A parabola.

Ex. 5. Through a fixed point O any line OP is drawn to meet a fixed line; to find the envelope of PQ drawn so as to make the angle OPQ constant.

Let OP make the angle 0 with the perpendicular on the fixed line, and its length is p sec0; but the perpendicular from O on PQ makes a fixed angle ẞ with OP, therefore its length is = p sec cos ß; and since this perpendicular makes an angle = 0 +ẞ with the perpendicular on the fixed line, if we assume the latter for the axis of x the equation of PQ is

or

x cos (0+ẞ) + y sin (8 + ß) = p sec0 cos ß,

x cos (20+ ẞ) + y sin (20 + ẞ) = 2p cosẞ - x cosß — y sin ß,

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x2 + y2= (x cosẞ+ y sin ẞ - 2p cos ẞ)2,

the equation of a parabola having the point O for its focus.

Ex. 6. Find the envelope of the line + = 1, where the indeterminates are

connected by the relation μ + μ' = C.

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We may substitute for ', C-μ, and clear of fractions; the envelope is thus found to be A2 + B2 + C2 - 2AB-2AC - 2BC = 0,

an equation to which the following form will be found to be equivalent,

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Thus, for example,-Given vertical angle and sum of sides of a triangle, to find the envelope of base.

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In like manner,-Given in position two conjugate diameters of an ellipse, and the sum of their squares, to find its envelope.

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The ellipse, therefore, must always touch four fixed right lines.

285. If the coefficients in the equation of any right line λa+μB+vy be connected by any relation of the second order in X, μg Vg

Αλ + Β' + v* + 2 Fμν + 2 ανλ + 21λμ = 0,

the envelope of the line is a conic section. Eliminating v between the equation of the right line and the given relations, we have (Ay2 — 2 Gya + Ca2) λ2 + 2 (Hy2 − Fya – Gyß + Caß) λμ

and the envelope is

+ (By2 − 2Ęyß + CB2) μ2 = 0,

(Ay2-2 Gya+Ca2) (By3-2Fyẞ+CB2)=(Hy"- Fyα- Gyẞ+Caß)2. Expanding this equation, and dividing by y2, we get

2

(BC − F2) a2 + (CA − G3) ẞ2 + (AB — H3) y2

+2(GH-AF) By + 2 (HF- BG) ya + 2 (FG - CH) aß = 0. · The result of this article may be stated thus: Any tangential equation of the second order in λ, μ, v represents a conic, whose trilinear equation is found from the tangential by exactly the same process that the tangential is found from the trilinear.

For it is proved (as in Art. 151) that the condition that λα + β + yy shall touch

aa2 + bẞ2 + cy2+ 2ƒBy + 2gya + 2haß =0,

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