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be regarded as a tangent, the point of contact of which is given (at an infinite distance). To have given the direction of an asymptote is equivalent to having one point given, for this virtually determines the point in which the conic meets the line at infinity.

9. If it be given that three given points form a conjugate triad, this is equivalent to three conditions, as the equation of the conic, when these are taken as angular points of the triangle of reference, is of the form

ua2 + vß2 + wy2 = 0.

Two more conditions will therefore completely determine the conic. If these conditions be that the conic shall pass through two given points, or touch two given straight lines, or pass through one given point and touch one given straight line, one conic only can be drawn to satisfy these conditions.

We may observe that, if the above conic pass through the point (f, g, h), it also passes through the three points (-f h), (f, g, h), (f, g, h), and that, if it touch 9, the line (l, m, n), it also touches the lines (− l, m, n), (l, — m, n), (l, m, — n).

ON THE LOCUS OF THE CENTRE OF A SYSTEM OF CONICS WHICH SATISFY FOUR CONDITIONS, EXPRESSED BY PASSING THROUGH POINTS AND TOUCHING STRAIGHT LINES.

10. The locus of the centre of a conic, which passes through m points, and touches n straight lines, m+n being equal to four, will be a conic, in every case except two. We will consider the several cases in order.

11. Let the system pass through four points.

This is best treated by Cartesian co-ordinates.

Of the conics which can be described passing through four points, two are parabolas. Take, as co-ordinate axes, that diameter of each of these parabolas, the tangent at the ex

tremity of which is parallel to the axis of the other. Then the two parabolas will be represented by the equations

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The system of conics is represented by the equation
x2 +λy2 + 2λg'x + 2fy + h + Xh′ = 0.........................(3),

A being an arbitrary multiplier.

The centre is given by the equations

x +λg'′ = 0, λy+fƒ=0.

Eliminating A, we get for the locus of the centres

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a conic, whose asymptotes are parallel to the axes of the parabolas (1) and (2).

If the four points form a convex quadrangle, the parabolas will be real, and the locus (4) an hyperbola. If the quadrangle be concave, the parabolas will be imaginary, and the locus of the centres an ellipse.

The curve (4) bisects the distance between each pair of the four points, and passes through the vertices of the quadrangle. This may be seen from geometrical considerations, for the three pairs of straight lines which belong to this system of conics, the vertices are respectively the centres.

From the form of the equation (3) we see that every conic of the system has a pair of conjugate diameters parallel to the axes of the parabola (1), (2); in other words to the asymptotes of (4).

The conic of minimum eccentricity is obtained by making λ = 1. In this case, these are the equal conjugate diameters.

If the axes of the parabolas be at right angles to one another, the four points lie on the circumference of a circle. The axes of every conic in (3) are then parallel to the coordinate axes, and (4) is a rectangular hyperbola.

If each of the four points be the orthocentre of the other

three, the system of conics is a system of rectangular hyperbolas, and (4) is the nine-point circle of the given points.

12. Let three points and a tangent be given.

In this case we may see, à priori, that the locus will be a curve of the fourth degree, for we can describe four parabolas satisfying the given conditions, and the locus will have four asymptotes, parallel to the axes of these parabolas.

Take the triangle formed by the three points for the triangle of reference, and use triangular co-ordinates. Let the tangent be represented by lx + my +nz =

= 0.

Then, if the system of conics be represented by

Xyz + μzx+vxy = 0,

the condition of tangency gives

2

l2x2 + m2μ2 + n2v2 – 2mnμv – 2nlvλ — 2lmλμ = 0. The centre is given by the equation

μz + vy = vx + λz = λy + μx.

If each member of this be put for the moment

=

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P, we have

2

p;

therefore the equation of the locus becomes

l2x2 (y + z−x)2 + m2y2 (z + x − y)2 + n2z2 (x + y − z)2

- 2mnyz (z + x − y) (x + y − z) — 2nl lx (x + y − z ) (y + z − x)

-

2lmxy (y + z- x) (z + x − y) = 0,

a curve of the fourth degree.

(z+x

Writing 1-2, 1-2y, 1–2z, for

y + z−x, 2 + x − y, x + y — z,

respectively, the terms of the fourth order become

l2x2 + m2y2 + n31⁄23 — 2mny31⁄23 — 2nlz*x* — 2lmx3y2.

Hence the asymptotes, and therefore the axes of the four parabolas, are parallel to the four lines

± 13 x ± m3y ± n3 z = 0.

13. Let two points and two tangents be given.

In this case, again, four parabolas can be described satisfying the given conditions, and we might therefore expect that the locus would be a curve of the fourth degree. It will be found, however, that it breaks up into two factors of the second degree.

Taking the line joining the two points as a = 0, and the other two as ẞ= 0, y = 0, the equation of the system may be written

2By + (λa + mß + ny)2 = 0.

Here is a variable parameter. For the determination of m and n, we may proceed as follows. Let the values of corresponding to a = 0, be called I, l'.

We have then

B

Υ

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For the centre, we have the equations

λ (λa + mẞ + ny) = m (λa +, mB + ny) + y

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(m + n) (λa + mB + ny) + B + y m + n .. λ= 2 (λa+mB+ny)

=

+

2

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n B + Y B-Y

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(m2 — mn − 1) B2 + (n2 — mn − 1) y2 — (m2 — 2mn + n2 — 2) By

+(n2 — mn) ya + (m2 — mn) aß = 0,

or, dividing by m2, and substituting the values of

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+ ll'ya+aß ± (ll')* (ya+aß) = 0,

giving, as stated above, two conic sections for the required locus.

14. Let one point, and three tangents be given.

Here the required locus will be a conic, since only two parabolas can be described satisfying the given conditions.

Take the three straight lines as lines of reference; and let f, g, h, be the triangular co-ordinates of the given point. We have then, as the equation of the system,

l2x2 + m2y2 + n2x2 — 2mn yz — 2nl zx — 2lm xy = 0,

subject to the condition

l2ƒ2 + m2g2 + n3h2 — 2mn gh — 2nl hf - 2lm fg = 0................(1).

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