or finally it is Ax + By + C = 0, showing that the point Q lies in a line the position of which is independent of the particular lines OAA′, OBB′ used in the construction. It is proper to notice that there is no correspondence to each other of the points A, A′ and B, B′; the grouping might as well have been A, A′ and B′, B; and it thence appears that the line Ax + By + C = 0 just obtained is in fact the line joining the point Q with the point R which is the intersection of AB and A′B′.
15. In § 8 it has been seen that two conditions determine the equation of a straight line, because in Ax + By + C = 0 one of the coefficients may be divided out, leaving only two parameters to be determined. Similarly five conditions instead of six determine an equation of the second degree (a, b, c, f, g, h)(x, y, 1)2 = 0, and nine instead of ten determine a cubic (*)(x, y, 1)3 = 0. It thus appears that a cubic can be made to pass through 9 given points, and that the cubic so passing through 9 given points is completely determined. There is, however, a remarkable exception. Considering two given cubic curves S = 0, S′ = 0, these intersect in 9 points, and through these 9 points we have the whole series of cubics S − kS′ = 0, where k is an arbitrary constant: k may be determined so that the cubic shall pass through a given tenth point (k = S0 ÷ S′0, if the coordinates are (x0, y0), and S0, S′0 denote the corresponding values of S, S′). The resulting curve SS′0 − S′S0 = 0 may be regarded as the cubic determined by the conditions of passing through 8 of the 9 points and through the given point (x0, y0); and from the equation it thence appears that the curve passes through the remaining one of the 9 points. In other words, we thus have the theorem, any cubic curve which passes through 8 of the 9 intersections of two given cubic curves passes through the 9th intersection.
The applications of this theorem are very numerous; for instance, we derive from it Pascal’s theorem of the inscribed hexagon. Consider a hexagon inscribed in a conic. The three alternate sides constitute a cubic, and the other three alternate sides another cubic. The cubics intersect in 9 points, being the 6 vertices of the hexagon, and the 3 Pascalian points, or intersections of the pairs of opposite sides of the hexagon. Drawing a line through two of the Pascalian points, the conic and this line constitute a cubic passing through 8 of the 9 points of intersection, and it therefore passes through the remaining point of intersection—that is, the third Pascalian point; and since obviously this does not lie on the conic, it must lie on the line—that is, we have the theorem that the three Pascalian points (or points of intersection of the pairs of opposite sides) lie on a line.
16. Metrical Theory resumed. Projections and Perpendiculars.—It is a metrical fact of fundamental importance, already used in § 8, that, if a finite line PQ be projected on any other line OO′ by perpendiculars PP′, QQ′ to OO′, the length of the projection P′Q′ is equal to that of PQ multiplied by the cosine of the acute angle between the two lines. Also the algebraical sum of the projections of the sides of any closed polygon upon any line is zero, because as a point goes round the polygon, from any vertex A to A again, the point which is its projection on the line passes from A′ the projection of A to A′ again, i.e. traverses equal distances along the line in positive and negative senses. If we consider the polygon as consisting of two broken lines, each extending from the same initial to the same terminal point, the sum of the projections of the lines which compose the one is equal, in sign and magnitude, to the sum of the projections of the lines composing the other. Observe that the projection on a line of a length perpendicular to the line is zero.
Let us hence find the equation of a straight line such that the perpendicular OD on it from the origin is of length ρ taken as positive, and is inclined to the axis of x at an angle xOD = α, measured counter-clockwise from Ox. Take any point P(x, y) on the line, and construct OM and MP as in fig. 48. The sum of the projections of OM and MP on OD is OD itself; and this gives the equation of the line
Observe that cos α and sin α here are the sin α and −cos α, or the −sin α and cos α of § 8 according to circumstances.
We can write down an expression for the perpendicular distance from this line of any point (x′, y ′) which does not lie upon it. If the parallel through (x′, y ′) to the line meet OD in E, we have x′ cos α + y ′ sin α = OE, and the perpendicular distance required is OD − OE, i.e. ρ − x′ cos α − y ′ sin α; it is the perpendicular distance taken positively or negatively according as (x′, y ′) lies on the same side of the line as the origin or not.
The general equation Ax + By + C = 0 may be given the form x cos α + y sin α − ρ = 0 by dividing it by √(A2 + B3). Thus (Ax′ + By ′ + C) ÷ √(A2 + B2) is in absolute value the perpendicular distance of (x′, y ′) from the line Ax + By + C = 0. Remember, however, that there is an essential ambiguity of sign attached to a square root. The expression found gives the distance taken positively when (x′, y ′) is on the origin side of the line, if the sign of C is given to √(A2 + B2).
17. Transformation of Coordinates.—We often need to adopt new axes of reference in place of old ones; and the above principle of projections readily expresses the old coordinates of any point in terms of the new.
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| Fig. 53. |
Suppose, for instance, that we want to take for new origin the point O′ of old coordinates OA = h, AO′ = k, and for new axes of X and Y lines through O′ obtained by rotating parallels to the old axes of x and y through an angle θ counter-clockwise. Construct (fig. 53) the old and new coordinates of any point P. Expressing that the projections, first on the old axis of x and secondly on the old axis of y, of OP are equal to the sums of the projections, on those axes respectively, of the parts of the broken line OO′M′P, we obtain:
and
Be careful to observe that these formulae do not apply to every conceivable change of reference from one set of rectangular axes to another. It might have been required to take O′X, O′Y′ for the positive directions of the new axes, so that the change of directions of the axes could not be effected by rotation. We must then write −Y for Y in the above.
Were the new axes oblique, making angles α, β respectively with the old axis of x, and so inclined at the angle β − α, the same method would give the formulae
18. The Conic Sections.—The conics, as they are now called, were at first defined as curves of intersection of planes and a cone; but Apollonius substituted a definition free from reference to space of three dimensions. This, in effect, is that a conic is the locus of a point the distance of which from a given point, called the focus, has a given ratio to its distance from a given line, called the directrix (see Conic Section). If e : 1 is the ratio, e is called the eccentricity. The distances are considered signless.
Take (h, k) for the focus, and x cos α + y sin α − p = 0 for the directrix. The absolute values of √{(x − h)2 + (y − k)2} and p − x cos α − y sin α are to have the ratio e : 1; and this gives
as the general equation, in rectangular coordinates, of a conic.
It is of the second degree, and is the general equation of that degree. If, in fact, we multiply it by an unknown λ, we can, by solving six simultaneous equations in the six unknowns λ, h, k, e, p, α, so choose values for these as to make the coefficients in the equation equal to those in any equation of the second degree which may be given. There is no failure of this statement in the special case when the given equation represents two straight lines, as in § 10, but there is speciality: if the two lines intersect, the intersection and either bisector of the angle between them are a focus and directrix; if they are united in one line, any point on the line and a perpendicular to it through the point are: if they are parallel, the case is a limiting one in which e and h2 + k2 have become infinite while e−2(h2 + k2) remains finite. In the case (§ 9) of an equation such as represents a circle there is another instance of proceeding to a limit: e has to become 0, while ep remains finite: moreover α is indeterminate. The centre of a circle is its focus, and its directrix has gone to infinity, having no special direction. This last fact illustrates the necessity, which is also forced on plane geometry by three-dimensional considerations, of treating all points at infinity in a plane as lying on a single straight line.
Sometimes, in reducing an equation to the above focus and directrix form, we find for h, k, e, p, tan α, or some of them, only imaginary values, as quadratic equations have to be solved; and we have in fact to contemplate the existence of entirely imaginary conics. For instance, no real values of x and y satisfy x2 + 2y2 + 3 = 0. Even when the locus represented is real, we obtain, as a rule, four sets of values of h, k, e, p, of which two sets are imaginary; a real conic has, besides two real foci and corresponding directrices, two others that are imaginary.
In oblique as well as rectangular coordinates equations of the second degree represent conics.
19. The three Species of Conics.—A real conic, which does not degenerate into straight lines, is called an ellipse, parabola or hyperbola according as e <, = , or > 1. To trace the three forms it is best so to choose the axes of reference as to simplify their equations.
In the case of a parabola, let 2c be the distance between the given focus and directrix, and take axes referred to which these are the point (c, 0) and the line x = − c. The equation becomes (x − c)2 + y2 = (x + c)2, i.e. y2 = 4cx.
In the other cases, take a such that a(e ~ e−1) is the distance of focus from directrix, and so choose axes that these are (ae, 0) and x = ae−1, thus getting the equation(x − ae)2 + y2 = e2(x − ae−1)2, i.e. (1 − e2)x2 + y2 = a2(1 − e2). When e < 1, i.e. in the case of an ellipse, this may be written x2/a2 + y2/b2 = 1, where b2 = a2(1 − e2); and when e > 1, i.e. in the case of an hyperbola, x2/a2 − y2/b2 = 1, where b2 = a2(e2 − 1).
