degree. We could, by the help of the same equation, find the locus of the intersection of tangents; if the sum, product, &c., of the intercepts on the axis be given.

Ex. 8. Given four tangents to a conic to find the locus of the centre. [The solution here given is by P. Serret, Nouvelles Annales, 2nd series, IV. 145.]

Take any axes, and let the equation of one of the tangents be x cos a + y sin a-p=0, then a is the angle the perpendicular on the tangent makes with the axis of x; and if be the unknown angle made with the same axis by the axis major of the conic, then a O is the angle made by the same perpendicular with the axis major. If then x and y be the co-ordinates of the centre, the formula of Art. 178 gives us

(x cos a + y sin a − p)2 = a2 cos2 (a — 0) + b2 sin2 (a — 0).

We have four equations of this

three unknown quantities a2, 62, 0.

form from which we have to eliminate the Using for shortness the abbreviation a for

x cos a + y sin a -p (Art. 53), this equation expanded may be written

a2 = (a2 cos20+b2 sin20) cos2a + 2 (a2 - b2) cos 0 sin @ cos a sin a + (a2 sin20+b2 cos20) sin2a. It appears then that the three quantities a2 cos20+ b2 sin2 0, (a2 - b2) cos e sin 0, a2 sin20+ b2 cos20, may be eliminated linearly from the four equations; and the result comes out in the form of a determinant

which expanded is of the form Aa2 + Bß2 + Cy2 + Dô2 = 0, where A, B, C, D are known constants. But this equation though apparently of the second degree is in reality only of the first; for if, before expanding the determinant, we write a2, &c., at full length, the coefficients of x2 are cos2 a, cos2 ß, cos2 y, cos2d; but these being the same as one column of the determinant, the part multiplied by a2 vanishes on expansion. Similarly, the coefficients of the terms xy and y2 vanish. The locus is therefore a right line. The geometrical determination of the line depends on principles to be proved afterwards: namely, that the polar of any point with regard to the conic is

Aa'a + BB'ß+ Cy'y + Dò'ò = 0;

and therefore that the polar of the point aß passes through yd. But when a conic reduces to a line by the vanishing of the three highest terms in its equation, the polar of any point is a parallel line at double the distance from the point. Thus it is seen that the line represented by the equation bisects the lines joining the points aß, yo; ay, Bo; ad, By. Conversely, if we are given in any form the equations of four lines a = 0, &c., the equation of the line joining the three middle points of diagonals of the quadrilateral may, in practice, be most easily formed by determining the constants so that Aa2 + Bẞ2 + Cy2 + Dô2 = 0 shall represent a right line.

Ex. 9. Given three tangents to a conic and the sum of the squares of the axes, find the locus of the centre. We have three equations as in the last example, and a fourth a2 + b2 = k2, which may be written

k2 = (a2 cos 0 + b2 sin20) + (a2 sin2 0 + b2 cos20),

and, as before, the result appears in the form of a determinant

which expanded is of the form Aa2 + Bß2 + Cy2 + D = 0. It is seen, as in the last example, that the coefficient of xy vanishes in the expansion, and that the coefficients of x2 and y2 are the same. The locus is therefore a circle. Now if Aa2+ Bẞ2 + Cy2=0 represents a circle, it will afterwards appear that the centre is the intersection of perpendiculars of the triangle formed by the lines a, ẞ, y. The present equation therefore, which differs from this by a constant, (Art. 81) represents a circle whose centre is the intersection of perpendiculars of the triangle formed by the three tangents.

Ex. 10. Given four points on a conic to find the locus of either focus. The distance of one of the given points from the focus (see Ex., p. 173) satisfies the equation p = Ax' + By + C.

We have four such equations from which we can linearly eliminate A, B, C, and we get the determinant ρ x' y' 1

which expanded is of the form lp + mp' + np" + pp"" = 0. If we look to the actual values of the coefficients l, m, n, p, and their geometric meaning (Art. 36), this equation geometrically interpreted gives us a theorem of Möbius, viz.

OA.BCD + OC.ABD OB. ACD + OD. ABC,

where O is the focus, and BCD the area of the triangle formed by three of the points (compare Art. 94). It is seen thus that l+m+ n + p = 0. If we substitute for p its value √(x-x′)2 + (y — y′)2}, &c., and clear of radicals, the equation of the locus is found to be of the sixth degree.

If the four given points be on a circle, Mr. Sylvester has remarked that the locus breaks up into two of the third degree, as Mr. Burnside has thus shewn. We have by a theorem of Feuerbach's, given Art. 94,

229. It is always advantageous to express the position of a point on a curve, if possible, by a single independent variable, rather than by the two co-ordinates x'y'. We shall, therefore, find it useful, in discussing properties of the ellipse, to make a substitution similar to that employed (Art. 102) in the case of the circle; and shall write

* The use of this angle was recommended by Mr. O'Brien, Cambridge Mathematical Journal, vol. IV., p. 99.

The geometric meaning of the angle 4 is easily explained.

If we describe a circle on the axis major as diameter, and produce the ordinate at P to meet the circle at Q, then the angle

QCL=0, for CL=CQ cos Q CL, or x'=a cos; and PL (Art. 163); or, since QLa sino, we have y'b sino.

=

b

a

QL

230. If we draw through P a parallel PN to the radius CQ, then PM::: PL: QL::b: a,

but

CQ=a, therefore PM=b.

PN parallel to CQ is, of course, = a.

Hence, if from any point of an ellipse A a line = a be inflected to the minor axis, its intercept to the axis major =b. If the ordinate PQ were produced to meet the circle again in the point Q', it could

D

L A

M

N

Β'

D'

be proved in like manner, that a parallel through P to the radius CQ' is cut into parts of a constant length. Hence, conversely, if a line MN, of a constant length, move about in the legs of a right angle, and a point P be taken so that MP may be constant, the locus of P is an ellipse, whose axes are equal to MP and NP. (See Ex. 12, p. 47).

On this principle has been constructed an instrument for describing an ellipse by continued motion, called the Elliptic Compasses. CA, CD' are two fixed rulers, MN a third ruler of a constant length, capable of sliding up and down between them, then a pencil fixed at any point of MN will describe an ellipse.

If the pencil be fixed at the middle point of MN it will describe a circle. (O'Brien's Co-ordinate Geometry, p. 112).

231. The consideration of the angle & affords a simple method of constructing geometrically the diameter conjugate to a

given one, for

Hence the relation

becomes

or

Q'

Hence we obtain the following construction. Let the ordinate at the given point P, when produced, meet the semicircle on the axis major at Q, join CQ, and erect CQ' perpendicular to it; then the perpendicular let fall on the axis from Q' will pass through P', a point on the conjugate diameter.

P'

с

M

M'

Hence, too, can easily be found the co-ordinates of P' given in Art. 172, for, since

x"

α

y"

=

y'

sing' =

coso, we have

=

b

a

From these values it appears that the areas of the triangles PCM, P' CM' are equal.

Ex. 1. To express the lengths of two conjugate semi-diameters in terms of the angle p. Ans. a'2 a2 cos2 + b2 sin2; b'2 = a2 sin2 + b2 cos2 p.

=

Ex. 2. To express the equation of any chord of the ellipse in terms of and o' (see p. 94).

Ex. 4. To express the length of the chord joining two points a, ß,

D2 = a2 (cosa - cos ß)2 + b2 (sin a — sin ß)2,

But (Ex. 1) the quantity between the parentheses is the semi-diameter conjugate to that to the point (a + ẞ); and (Ex. 2, 3) the tangent at the point (a + ẞ) is parallel to the chord joining the points a, ß; hence, if b' denote the length of the semidiameter parallel to the given chord, D= 26' sin (a – ẞ).

Ex. 5. To find the area of the triangle formed by three given points a, ß, y.
By Art. 36 we have

ΩΣ

ab {sin (a - ẞ) + sin (ẞ − y) + sin (y - a)}

--

= ab {2 sin ¦ (a — ẞ) cos 1 (a — ß) — 2 sin≥ (a — ß) cos } (a + ß − 2y)}

Ex. 6. If the bisectors of sides of an inscribed triangle meet in the centre, its area is constant.

Ex. 7. To find the radius of the circle circumscribing the triangle formed by three given points a, ß, y.

If d, e, f be the sides of the triangle formed by the three points,

where b', b', b'" are the semi-diameters parallel to the sides of the triangle. If c', c', c" be the parallel focal chords, then (see Ex. 5, p. 202) R2 =

c'c'"c"
4p

expressions are due to Mr. MacCullagh, Dublin Exam. Papers, 1836, p. 22.) Ex. 8. To find the equation of the circle circumscribing this triangle.

(These

= 1 (a2 + b2) − 1 (a2 − b2) {cos (a + ß) + cos (ß + y) + cos (y + α)}.

From this equation the co-ordinates of the centre of this circle are at once obtained. Ex. 9. The area of the triangle formed by three tangents is, by Art. 39,

ab tan (a - ẞ) tan (6 - y) tan (y-a).

Ex. 10. The area of the triangle formed by three normals is

tan (a — ß) tan 1 (6 − y) tan 1 (y − a) {sin (ß + y) + sin (y + a) + sin (a + ß)}2, 4ab consequently three normals meet in a point if

sin (ẞ + y) + sin (y + a) + sin (a + B) = 0. [Mr. Burnside.]

Ex. 11. To find the locus of the intersection of the focal radius vector FP with the radius of the circle CQ.

Let the central co-ordinates of P be x'y', of O, xy, then we have, from the similar triangles, FON, FPM,

y

=

y'

b sin o

x + c x2 + c a (e + cos)

Now, since is the angle made with the axis by the radius vector to the point 0, we at once obtain the polar equation of the locus by writing p cos o for x, p sind for y, and we find

Hence (Art. 193) the locus is an ellipse, of which C is one focus, and it can easily be proved that F is the other.

Ex. 12. The normal at P is produced to meet CQ; the locus of their intersection is a circle concentric with the ellipse.

but we may, as in the last example, write p cosp and p sin for x and y, and the equation becomes

(a - b) p = c2, or p = a + b.

Ex. 13. Prove that tan PFC =

tan 14.

+e

EE