These results may be written in another form, since 2 (y'x" - y′′x') = (x' + x') (y' -- y') − (y' + y′′) (x' — x"), Ex. 4. To find the co-ordinates of the intersection of the normals at the points x'y', x"y". where X, Y are the co-ordinates of the intersection of tangents, found in the last Example. The foci of an hyperbola are two points on the transverse axis, at a distance from the centre still = +c, c being in the hyperbola = √(a2 + b2). To express the distance of any point on an ellipse from the focus. square of the distance of any point from it = (x' − c)2 + y22 = x2 + y22 − 2cx' + c2. But (Art. 177) Hence = 0), the [We reject the value (ex – a) obtained by giving the other sign to the square root. For, since x is less than a, and e less than 1, the quantity ex - a is constantly negative, and, therefore, does not concern us, as we are now considering, not the direction, but the absolute magnitude of the radius vector FP.] Y We have, similarly, the distance from the other focus F'P = a + ex, since we have only to write - c for + c in the preceding formula. FP+ F'P 2a, Hence = or, The sum of the distances of any point on an ellipse from the foci is constant and equal to the axis major. 187. In applying the preceding proposition to the hyperbola, we obtain the same value for FP2; but in extracting the square root we must change the sign in the value of FP, for in the hyperbola a is greater than a, and e is greater than 1. Hence, a ex is constantly negative; the absolute magnitude, therefore, of the radius vector is Therefore, in the hyperbola, the difference of the focal radii is constant, and equal to the transverse axis. For both curves the rectangle under the focal radii = a2 – e2x2, that is (Art. 177), is equal to the square of the semiconjugate dia meter. 188. The reader may prove the converse of the above results by seeking the locus of the vertex of a triangle, if the base and either sum or difference of sides be given. Taking the middle point of the base (= 2c) for origin, the equation is √ {y2 + (c + x)2} ± √ {y2 + (c − x)2} = 2a, which, when cleared of radicals, becomes Now, if the sum of the sides be given, since the sum must always be greater than the base, a is greater than c, therefore the coefficient of y2 is positive, and the locus an ellipse. If the difference be given, a is less than c, the coefficient of y2 is negative, and the locus an hyperbola. 189. By the help of the preceding theorems, we can describe an ellipse or hyperbola mechanically. R If the extremities of a thread be fastened at two fixed points F and F', it is plain that a pencil moved about so as to keep the thread always stretched will describe an ellipse whose foci are F and F', and whose axis major is equal to the length of the thread. In order to describe an hyperbola, let a ruler be fastened at one extremity (F), and capable of moving round it, then if a thread, fastened to a fixed point F', and also to a fixed point on the ruler (R), be kept stretched by a ring at P, as the ruler is moved round, the point P will describe an hyperbola; for, since the sum of F'P and PR is constant, the difference of FP and F'P will be constant. F' 190. The polar of either focus is called the directrix of the conic section. The directrix must, therefore (Art. 173), be a line perpendicular to the axis major at a distance from the centre = a2 . C Knowing the distance of the directrix from the centre, we can find its distance from any point But the distance of any point on the curve from the focus = a - ex'. Hence we obtain the important property, that the distance of any point on the curve from the focus is in a constant ratio to its distance from the directrix, viz., as e to 1. Conversely, a conic section may be defined as the locus of a point whose distance from a fixed point (the focus) is in a constant ratio to its distance from a fixed line (the directrix). On this definition several writers have based the theory of conic sections. Taking the fixed line for the axis of x, the equation of the locus is at once written down which it is easy to see will represent an ellipse, hyperbola, or parabola, according as e is less, greater than, or equal to 1. Ex. If a curve be such that the distance of any point of it from a fixed point can be expressed as a rational function of the first degree of its co-ordinates, then the curve must be a conic section, and the fixed point its focus (see O'Brien's "Co-ordinate Geometry,” p. 85). For, if the distance can be expressed p = Ax+ By + C, since Ax + By + C is proportional to the perpendicular let fall on the right line whose equation is (Ax + By + C = 0), the equation signifies that the distance of any point of the curve from the fixed point is in a constant ratio to its distance from this line. 191. To find the length of the perpendicular from the focus on the tangent. The length of the perpendicular from the focus (+ c, 0) on the line ( (xxyy + = 1 is, by Art. 27, or, The rectangle under the focal perpendiculars on the tangent is constant, and equal to the square of the semiaxis minor. This property applies equally to the ellipse and the hyperbola. 192. Some important consequences may be drawn from the value of the perpendicular just found. Hence the sine of the angle which the focal radius vector We find, in like manner, the same value for sin F' PT, the sine of the angle which the other focal radius vector makes with the tangent. Hence the focal radii make equal angles with the tangent. This property is true both for the ellipse and hyperbola, and, on looking at the figures, it is evident that the tangent to the ellipse is the external bisector of the angle between the focal radii, and the tangent to the hyperbola the internal bisector. Hence, if an ellipse and hyperbola, having the same foci, pass through the F T' P same point, they will cut each other at right angles, that is to say, the tangent to the ellipse at that point will be at right angles to the tangent to the hyperbola. Ex. 1. Prove analytically that confocal conics cut at right angles. But if the conics be confocal a2 — a'2 = b2 — b'2, and this relation becomes Ex. 2. Find the length of a line drawn through the centre parallel to either focal radius vector, and terminated by the tangent. ab This length is found by dividing the perpendicular from the centre on the tangent by the sine of the angle between the radius vector and tangent therefore α. (C) and is 193. The normal, being perpendicular to the tangent, is the internal bisector of the angle between the focal radii in the case of the ellipse, and the external bisector in the case of the hyperbola. We can give an independent proof of this, by showing that |