Ex. 10. If normals be drawn at the extremities of any focal chord, a line drawn through their intersection parallel to the axis major will bisect the chord. [This solution is by Larrose, Nouvelles Annales XIX. 85].

Since each normal bisects the angle between the focal radii, the intersection of normals at the extremities of a focal chord is the centre of the circle inscribed in the triangle whose base is that chord, and sides the lines joining its extremities to the other focus. Now if a, b, c be the sides of a triangle whose vertices are x'y', x"y", x""y"", then, Ex. 6, p. 6, the co-ordinates of the centre of the inscribed circle are

In the present case the co-ordinates of the vertices are x', y'; x", y'; — c, 0; and the lengths of opposite sides are a + ex'', a + ex', 2a - ex' — ex". We have therefore

or, reducing by the first relation of the last Example, y = 1 (y' + y'), which proves the theorem.

We could find, similarly, expressions for the co-ordinates of the intersection of tangents at the extremities of a focal chord, since this point is the centre of the circle exscribed to the base of the triangle just considered. The line joining the intersection of tangents to the corresponding intersection of normals evidently passes through a focus, being the bisector of the vertical angle of the same triangle.

Ex. 11. To find the locus of the intersection of normals at the extremities of a focal chord.

Let a, ẞ be the co-ordinates of the middle point of the chord, and we have, by the last Example,

If, then, we knew the equation of the locus described by aß, we should by making the above substitutions have the equation of the locus described by xy. Now the polar equation of the locus of middle point, the focus being origin, is (Art. 193)

which transformed to rectangular axes, the centre being origin, becomes

b2a2 + a2ß2 = b2ca.

The equation of the locus sought is, therefore,

a2b2 (x + c)2 + (a2 + c2)2y2 = b2c (a2 + c2) (x + c).

Ex. 12. If be the angle between the tangents to an ellipse from any point P and if p, p' be the distances of that point from the

p2 + p12 - 4a2
2pp'

sin TPF.sintPF=

FT.F'T' b2
PF.PF' φρ

T

T

Ex. 13. If from any point O two lines be drawn to the foci (or touching any confocal conic) meeting the conic in R, R'; S, S'; then

It appears from the quadratic, by which the radius vector is determined (Art. 136), that the difference of the reciprocals of the roots will be the same for two values of 0, which give the same value to

(ac - g2) cos2 0 + 2 (ch − gf) cos 0 sin 0 + (bc − ƒ2) sin2 0.

Now it is easy to see that A cos20 + 2H cos 0 sin 0 + B sin2 0 has equal values for any two values of 0, which correspond to the directions of lines equally inclined to the two represented by Ax2 + 2Hxy + By2 = 0. But the function we are considering becomes = 0 for the direction of the two tangents through O (Art. 147) and tangents to any confocal are equally inclined to these tangents (Art. 189). It follows from this example that chords which touch a confocal conic are proportional to the squares of the parallel diameters (see Ex. 15, p. 210).

227. We give in this Article some examples on the parabola. The reader will have no difficulty in distinguishing those of the examples of the last Article, the proofs of which apply equally to the parabola.

Ex. 1. Find the co-ordinates of the intersection of the two tangents at the points x'y', x'y', to the parabola y2 = px. y' + y" Ans. y=

y'y"

x=

Ρ

Ex. 2. Find the locus of the intersection of the perpendicular from focus on tangent with the radius vector from vertex to the point of contact.

Ex. 3. The three perpendiculars of the triangle formed by three tangents intersect on the directrix (Steiner, Gergonne, Annales, XIX. 59; Walton, p. 119).

The equation of one of those perpendiculars is (Art. 32)

y'y"" - y'y" y"y"" _y" — y'"' ( y − '' + y'") = '

Р

x

+

+ (x+2)

Ρ

2

= 0;

= 0.

The symmetry of the equation shows that the three perpendiculars intersect on the directrix at a height

Ex. 4. The area of the triangle formed by three tangents is half that of the triangle formed by joining their points of contact (Gregory, Cambridge Journal, II. 16 ; Walton, p. 137. See also Lessons on Higher Algebra, Ex. 12, p. 14).

Ꭰ Ꭰ

Substituting the co-ordinates of the vertices of the triangles in the expression of

1

Art. 36, we find for the latter area, (y' — y'') (y" — y''') (y"" — y'); and for the former area half this quantity.

2p

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

But if d be the length of the

def d, e, f, and whose area = Σ is easily proved to be 42 chord joining the points x"y", ""y"", and e' the angle which this chord makes with the axis, it is obvious that d sin 0' y" - y"". Using, then, the expression for the Р area found in the last Example, we have R We might ex2 sin e' sin 0" sin 0"" 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 with the axis

c'c'c'"
4p

=

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 0" sin 0""

= ; or R2:

P'p'p""
"
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

(y'2 - 4mx') (y2 — 4mx) = {yy' — 2m (x + x')}2.

A parallel pair of lines through the origin is

x'y2 - y'xy + mx2 = 0.

The angle contained by which is (Art. 74) tan o

√(y'2 - 4mx') = x' + m

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 tan2 4, or y2 + (x − m)2 = (x + m)2 sec2 p. 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 co-ordinates of the intersection of the normals at the points x'y', x"y".

y'2 + y'y" + y'

Ans. x 2m+

4m

y=

y'y" (y+y") 8m2

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

B2

αβ

x = 2m+

α,

y =

m

m

Ex. 11. Find the co-ordinates 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+Y2+Y3 = 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. In this case a=- m, x = 3m+

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

length of the perpendicular on the chord from the intersection of the tangents). But 2wy = 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)

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. 143)

μμ'χ2 + 2hry + λλ' - μμ' (λ + λ') α - λλ' (μ + μ') y + λλ'μμ' = 0.

And if the axes be rectangular, this will represent an equilateral hyperbola (Art. 174) if XX' = — μμ'. If, therefore, the axes be any side of the given triangle, and the perpendicular on it from the opposite vertex, the portions X, X', μ 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.

μ

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. 148). Ex. 3. A conic being given by the general equation, find the condition that the pole of the axis of x should lie on the axis of y, and vice versa. Ans. hc=fg.

Ex. 4. In the same case, what is the condition that an asymptote should pass through the origin? Ans. af2-2fgh + 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 w + y2) +fx + 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 co-ordinates of the centre, provided we have a + b = 2h cosw, 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