280*. The equation (Art. 258 (a)) x2+ y2=e3y* (where the origin is a focus and y the corresponding directrix) is a particular case of that just considered. The tangents through (y, x) to the curve are evidently ey +x and ey-x. If, therefore, the curve be a parabola, e= 1; and the tangents are the internal and external bisectors of the angle (yx). Hence, "tangents to a parabola from any point on the directrix are at right angles to each other."

In general, since x = ey cosp, y=ey sin p, we have

or expresses the angle which any radius vector makes with x. Hence we can find the envelope of a chord which subtends a constant angle at the focus, for the chord.

x cos (+☀') + y sin (p + ☀') = ey cus 1 (4 − 4'),

if - ' be constant, must, by the present section, always touch x2+ y2=e2y2 cos' ('),

a conic having the same focus and directrix as the given one.

281. The line joining the focus to the intersection of two tangents is found by subtracting

to be

x cos y sin ey=0,

x cos p' + y sin p' — ey = 0,

x sin(+)- y cos (p + p')=0,

the equation of a line making an angle (+$′) with the axis of x, and therefore bisecting the angle between the focal radii. The line joining to the focus the point where the chord of contact meets the directrix is

x cos (+)+ y sin 1⁄2 (4 + 6') = 0,

a line evidently at right angles to the last.

To find the locus of the intersection of tangents at points which subtend a given angle 28 at the focus.

By an elimination precisely the same as that in Ex. 2, Art. 102, the equation of the locus is found to be (x + y2) cos2 d = e2y2,

*Art. 279 of the older editions now numbered Art. 258 (a).

which represents a conic having the same focus and directrix as

the given one, and whose eccentricity

=

=

e

COS δ

If the curve be a parabola, the angle between the tangents is in this case given. For the tangent (x cos + y sin – y) bisects the angle between x cos +y sin and y. The angle between the tangents is, therefore, half the angle between x cosp+y sino and x cos p' + y sin p', or (4-6). 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. 148) 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, B, y, the equations of the conics can be expressed in the form aa2+bB+cy2 = 0, 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 coordinates 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 x2+ y2 - 22, and the other ax2 + by2 — cz2. Then, as at Ex. 1, p. 94, the coordinates of the intersection of tangents at points a, 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 (ay);

and, since the coordinates of the point whose locus sin (a + B), cos} (a – ẞ), the equation of the locus is

— B) cos y,

(a — ẞ) sin y;

we seek are cos (a + ẞ),

a − b)2 − (a + b — c)2 •
− = —

Ex. 2. A triangle is inscribed in the conic x2 + y2 = 22; and two sides touch the conic ax2 + by2=cz2; find the envelope of the third side.

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

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 μL - 2μR+M, where is indeterminate, always touches the curve LM-R2. 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 μ+k 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

L L.

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

μ3L− 2μR + M=0, μL-R=0;

or, 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 (Art. 147), 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-R2.

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 cosp+ 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

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. 147); 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

μ2L'-2μR'+M' =0,

where L', R', M' are the results of substituting the coordinates 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 coordinates satisfy the equation LM-R2. 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 a + uß be the line joining to aß the vertex which moves along y, then the equations of the sides through the fixed points are

ỷ (a + MB) - (ẻ tuổi) y = 0, g” (a + MB) - (a + ußng 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′′y) (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 coordinates of the fixed points may be y = 0, x =±c; then if the variable line be y mx + n = 0, we have by the condition of the question

or

but

therefore

and the envelope is

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 + b2-202 b2

1.

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

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 with the perpendicular on the fixed line, and its length is p sec 0; 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

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