Substituting the coordinates of the vertices of the triangles in the expression of

Ex. 5. Find an expression for the radius of the circle circumscribing a triangle inscribed in a parabola.

The radius of the circle circumscribing a triangle, the lengths of whose sides are d, e, f, and whose area = Σ is easily proved to be But if d be the length of the

def 42

chord joining the points "y", ""y""', and ' the angle which this chord makes with the axis, it is obvious that d sin 0′y" - y"". Using, then, the expression for the p We might ex2 sin 0'sin " sin ""*

area found in the last Example, we have R= press the radius, also, in terms of the focal chords parallel to the sides of the triangle. For (Art. 193, Ex. 2) the length of a chord making an angle 0 with the axis

It follows from Art. 212 that c', c", c"" are the parameters of the diameters which bisect the sides of the triangle.

Ex. 6. Express the radius of the circle circumscribing the triangle formed by three tangents to a parabola in terms of the angles which they make with the axis.

p

8 sin 0' sin e" sin ""; or R2:

64p

Ans. R= where p', p", p"" are the parameters of the diameters through the points of contact of the tangents (see Art. 212).

Ex. 7. Find the angle contained by the two tangents through the point x'y' to the parabola y2 = 4mx.

The equation of the pair of tangents is (as in Art. 92) found to be

Ex. 8. Find the locus of intersection of tangents to a parabola which cut at a given angle.

Ans. The hyperbola, y2 - 4mx = (x + m)2 tan2p, or y2 + (x — m)2 = (x + m)2 sec2p. From the latter form of the equation it is evident (see Art. 186) that the hyperbola has the same focus and directrix as the parabola, and that its eccentricity = sec p.

Ex. 9. Find the locus of the foot of the perpendicular from the focus of a parabola on the normal.

-

The length of the perpendicular from (m, 0) on 2m (y − y') + y' (x − x') = 0 is

But if 0 be the angle made with the axis by the perpendicular (Art. 212)

Ex. 10. Find the coordinates of the intersection of the normals at the points x'y', x"y".

Ans. x 2m+

y'2 + y'y" + y'′′

112

y=

4m

y'y" (y' + y′′) 8m2

Or if a, ẞ be the coordinates of the corresponding intersection of tangents, then (Ex. 1)

Ex. 11. Find the coordinates of the points on the curve, the normals at which pass through a given point x'y'.

Solving between the equation of the normal and that of the curve, we find

2y3 + (p2 - 2px') y = p2y',

and the three roots are connected by the relation y1 + 1⁄2 + 1⁄2 = 0. The geometric meaning of this is, that the chord joining any two, and the line joining the third to the vertex, make equal angles with the axis.

Ex. 12. Find the locus of the intersection of normals at the extremities of chords which pass through a given point x'y'.

We have then the relation ẞy' = 2m (x' + a); and on substituting in the results of Ex. 10 the value of a derived from this relation we have

2mx+ By' = 4m2 + 2ß2 + 2mx'; 2m2y = 2ẞmx' — ẞ2y' ;

whence, eliminating ß, we find

2 {2m (y - y') + y′ (x − x')}2 = (4mx′ — y'2) (y'y + 2x'x — 4mx′ — 2x'2),

the equation of a parabola whose axis is perpendicular to the polar of the given point. If the chords be parallel to a fixed line, the locus reduces to a right line, as is also evident from Ex. 11.

Ex. 13. Find the locus of the intersection of normals at right angles to each other.

Ex. 14. If the lengths of two tangents be a, b, and the angle between them w, find the parameter.

Draw the diameter bisecting the chord of contact; then the parameter of that

y2 sin20 w2y2

= (where is the 42-3

length of the perpendicular on the chord from the intersection of the tangents). But 2ay = ab sin w, and 16x2 = a2 + b2 + 2ab cos w; hence

Ex. 15. Show, from the equation of the circle circumscribing three tangents to a parabola, that it passes through the focus.

The equation of the circle circumscribing a triangle being (Art. 124)

By sin A+ ya sin B + aß sin C = 0;

the absolute term in this equation is found (by writing at full length for a, x cos a + y sin a -p, &c.) to be p'p" sin (ẞ- y) + p'p sin (y- a) + pp' sin (a — ẞ). But if the line a be a tangent to a parabola, and the origin the focus, we have (Art. 219) and the absolute term

=

m2

(sin (By) cosa + sin (y - a) cos ẞ + sin (a - ẞ) cos y}, cos a cos ẞ cos y which vanishes identically.

Ex. 16. Find the locus of the intersection of tangents to a parabola, being given either (1) the product of sines, (2) the product of tangents, (3) the sum or (4) difference of cotangents of the angles they make with the axis.

Ans. (1) a circle, (2) a right line, (3) a right line, (4) a parabola,

228. We add a few miscellaneous examples.

Ex. 1. If an equilateral hyperbola circumscribe a triangle, it will also pass through the intersection of its perpendiculars (Brianchon and Poncelet; Gergonne, Annales, XI., 205; Walton, p. 283).

The equation of a conic meeting the axes in given points is (Ex. 1, p. 148)

μμ' 2 + 2hry + λλ' ο - μμ' (λ + λ') = - λλ' (μ + μ^) y + λλόμμ' = 0.

-

And if the axes be rectangular, this will represent an equilateral hyperbola (Art. 174) if XX' -'. If, therefore, the axes be any side of the given triangle, and the perpendicular on it from the opposite vertex, the portions A, λ', u are given, thereλλ' fore, u' is also given; or the curve meets the perpendicular in the fixed point y=-~ which is (Ex. 7, p. 27) the intersection of the perpendiculars of the triangle. Ex. 2. What is the locus of the centres of equilateral hyperbolas through three given points?

Ans. The circle through the middle points of sides (see Ex. 3, p. 153). Ex. 3. A conic being given by the general equation, find the condition that the pole of the axis of ≈ should lie on the axis of y, and vice versa. Ans. hcfg.

Ex. 4. In the same case, what is the condition that an asymptote should pass through the origin? Ans. af22fgh + bg2 = 0.

Ex. 5. The circle circumscribing a triangle, self-conjugate with regard to an equilateral hyperbola (see Art. 99), passes through the centre of the curve. (Brianchon and Poncelet; Gergonne, XI. 210; Walton, p. 304). [This is a particular case of the theorem that the six vertices of two self-conjugate triangles lie on a conic (see Ex. 1, Art. 375).]

The condition of Ex. 3 being fulfilled, the equation of a circle passing through the origin and through the pole of each axis is

or

h (x2 + 2xy cos ∞ + y2) +ƒx+gy = 0,

x (hx + by +f) + y (ax + hy + g) − (a + b − 2h cos w) xy,

an equation which will evidently be satisfied by the coordinates of the centre, provided we have a + b = 2h cos ∞, that is to say, provided the curve be an equilateral hyperbola (Arts. 74, 174).

Ex. 6. A circle described through the centre of an equilateral hyperbola, and through any two points, will also pass through the intersection of lines drawn through each of these points parallel to the polar of the other.

Ex. 7. Find the locus of the intersection of tangents which intercept a given length on a given fixed tangent.

The equation of the pair of tangents from a point x'y' to a conic given by the general equation is given Art. 92. Make y = 0, and we have a quadratic whose roots

are the intercepts on the axis of x.

Forming the difference of the roots of this equation, and putting it equal to a constant, we obtain the equation of the locus required, which will be in general of the fourth degree; but if g2 = ac, the axis of a will touch the given conic, and the equation of the locus will become divisible by y2, and will reduce to the second

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. solution here given is by P. Serret, Nouve les Annales, 2nd series, IV. 145.]

[The

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 is the angle made by the same perpendicular with the axis major. If then x and y be the coordinates of the centre, the formula of Art. 178 gives us

We have four equations of this form from which we have to eliminate the three unknown quantities a2, 62, 0. Using for shortness the abbreviation a for a cosa + y sin a -p (Art. 53), this equation expanded may be written

a2= (a2 cos20 + b2 sin20) cos2a + 2 (a2 — b2) cos✪ sine cosa sina + (a2 sin20+b2 cos20) sin2a. It appears then that the three quantities a2 cos20 + b2 sin20, (a2 — b2) cos 0 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+ Do2 = 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 cos2a, cos2ß, cos2y, cos2d; but these being the same as one column of the determinant, the part multiplied by 2 vanishes ou expansion. Similarly, the coefficients of the terms ry 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 + Eß'ß + Cy'y + Dò'd = 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, yô; ay, ßồ; aồ, ßy. 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 cos20+ b2 sin20) + (a2 sin20 + 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 2 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.

If we consider the case of the equilateral hyperbola a2 + b2 = 0, we see that two equilateral hyperbolas can be described to touch four given lines, the centres being the intersections of the line joining the middle points of diagonals with any one of four circles whose centres are the intersections of perpendiculars of the four triangles formed by any three of the four given lines. From the fact that the four circles have two common points it follows that the four intersections of perpendiculars lie on a right line, perpendicular to the line joining middle points of diagonals (see Art. 268, Ex. 2).

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., Art. 186) 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

P, x' y' 1

p', x", y", 1

"" ', y"", 1

p"", "', y'""', 1 = 0,

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 + 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, though apparently of the eighth, is found to be only of the sixth degree. In fact, we may clear of radicals by giving each radical its double sign, and multiplying together the eight factors lp±mp' ± np" ± pp""; and then it is apparent that the highest powers in x and y will be (x2 + y2) multiplied by the product of the factors l+m+n+p; and that these terms vanish in virtue of the relation l+m+n+ p = 0. 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,

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

FF.