having the same foci, pass through the 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. The co-ordinates of the intersection of the conics satisfy the relation obtained by subtracting the equations one from the other, viz. 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. This length is found by dividing the perpendicular from the centre on the tangent the sine of the angle between the radius vector and tangent, and is ab b' by (+), therefore = a. Ex. 3. Verify that the normal, which is a bisector of the angle between the focal radii, divides the distance between the foci into parts which are proportional to the focal radii (Euc. VI. 3). The distance of the foot of the normal from the centre is (Art. 180) = e2x'. Hence its distances from the foci are c+ e2x' and c - e2x', quantities which are evidently e times a + ex' and a- ex'. Ex. 4. To draw a normal to the ellipse from any point on the axis minor. Ans. The circle through the given point, and the two foci, will meet the curve at the point whence the normal is to be drawn. 189. Another important consequence may be deduced from the theorem of Art. 187, that the rectangle under the focal perpendiculars on the tangent is constant. For, if we take any two tangents, we have (see figure, next page) FT but is the ratio of the sines of the parts into which the line Ft F"t' FP divides the angle at P, and is the ratio of the sines of F"T the parts into which F"P divides the same angle; we have, therefore, the angle TPF=t'PF". T F P T' F' If we conceive a conic section to pass through P, having Fand F" for foci, it was proved in Art. 188, that the tangent to it must be equally inclined to the lines FP, F"P; it follows, therefore, from the present Article, that it must be also equally inclined to PT, Pt; hence we learn that if through any point (P) of a conic section we draw tamgents (PT, Pt) to a confocal conic section, these tangents will be equally inclined to the tangent at P. 190. To find the locus of the foot of the perpendicular let fall from either focus on the tangent. The perpendicular from the focus is expressed in terms of the angles it makes with the axis by putting x'=c, y' = 0 in the formula of Art. 178, viz., p = √(a cosa +b sin a) - x' cosa-y' sina. Hence the polar equation of the locus is or or p=√(a2 cos a+b2 sin3a) — c cosa, p2+2cp cosa + c2 cosa a cosa + b2 sin3a, This (Art. 95) is the polar equation of a circle whose centre is on the axis of x, at a distance from the focus c; the circle is, therefore, concentric with the curve. The radius of the circle is, by the same Article, a. Hence, If we describe a circle having for diameter the transverse axis of an ellipse or hyperbola, the perpendicular from the focus will meet the tangent on the circumference of this circle. Or, conversely, if from any point F (see figure, p. 171) we draw a radius vector FT to a given circle, and draw TP perpendicular to FT, the line TP will always touch a conic section, having Ffor its focus, which will be an ellipse or hyperbola, according as Fis within or without the circle. It may be inferred from Art. 188, Ex. 2, that the line CT, whose length = a, is parallel to the focal radius vector F"P. 191. To find the angle subtended at the focus by the tangent drawn to a central conic from any point (xy). Let the point of contact be (x'y'), the centre being the origin, then, if the focal radii to the points (xy), (x'y'), be p, p', and make angles 0, 0', with the axis, it is evident that but from the equation of the tangent we must have a* xx' + b2, =e2xx' + cx + cx' + a2 = (a + ex) (a+ex') ; ρ or, since p'=a+ex', we have, (see O'Brien's Co-ordinate Geometry, p. 156), cos (0-0): = a + ex Since this value depends solely on the co-ordinates xy, and does 192. The line joining the focus to the pole of any chord passing through it is perpendicular to that chord. This may be deduced as a particular case of the last Article, the angle subtended at the focus being in this case 180°; or directly as follows:-The equation of the perpendicular through any point x'y' to the polar of that point + =1) Art. 180, a2 'xx' yy 1) is, as in seers (over) But if x'y' be anywhere on the directrix, we have x' it will then be found that both the equation of the polar and that of the perpendicular are satisfied by the co-ordinates of the focus (x = c, y=0). A A When in any curve we use polar co-ordinates, the portion intercepted by the tangent on a perpendicular to the radius vector drawn through the pole is called the polar subtangent. Hence the theorem of this Article may be stated thus: The focus being the pole, the locus of the extremity of the polar subtangent is the directrix. It will be proved (Chap. XII.) that the theorems of this and the last Article are true also for the parabola. Ex. 1. The angle is constant which is subtended at the focus, by the portion intercepted on a variable tangent between two fixed tangents. By Art. 191, it is half the angle subtended by the chord of contact of the fixed tangents. Ex. 2. If any chord PP' cut the directrix in D, then FD is the external bisector of the angle PFP'. For FT is the internal bisector (Art. 191); but D is the pole of FT (since it is the intersection of PP', the polar of T, with the directrix, the polar of F); therefore, DF is perpendicular to FT, by 192 and is therefore the external bisector. [The following theorems (communica ted to me by the Rev. W. D. Sadleir) are D T F founded on the analogy between the equations of the polar and the tangent.] P' Ex. 3. If a point be taken anywhere on a fixed perpendicular to the axis, the perpendicular from it on its polar will pass through a fixed point on the axis. For the intercept made by the perpendicular will (as in Art. 180) be e2x', and will therefore be constant when x' is constant. Ex. 4. Find the lengths of the perpendicular from the centre and from the foci on the polar of x'y'. Ex. 5. Prove CM.PN' = b2. This is analogous to the theorem that the rectangle under the normal and the central perpendicular on P 193. To find the polar equation of the ellipse or hyperbola, the focus being the pole. The length of the focal radius vector (Art. 182) = a − ex' ; but a' (being measured from the centre) p cose + c. or = p=a-ep cos — ec, Hence The double ordinate at the focus is called the parameter; its half is found by making = 90° in the equation just given, to be = α = a(1— e3). The parameter is commonly denoted by the letter p. Hence the equation is often written The parameter is also called the Latus Rectum. Ex. 1. The harmonic mean between the segments of a focal chord is constant, and equal to the semi-parameter. For, if the radius vector FP, when produced backwards through the focus, meet Ex. 2. The rectangle under the segments of a focal chord is proportional to the whole chord. This is merely another way of stating the result of the last Example; but it may be proved directly by calculating the quantities FP. FP', FP + FP', which are easily seen to be, respectively Ex. 3. Any focal chord is a third proportional to the transverse axis and the parallel diameter. For it will be remembered that the length of a semi-diameter making an angle with the transverse axis is (Art. 161) R2= b2 2R2 Hence the length of the chord FP+ FP' found in the last Example = α Ex. 4. The sum of two focal chords drawn parallel to two conjugate diameters is constant. For the sum of the squares of two conjugate diameters is constant (Art. 173). Ex. 5. The sum of the reciprocals of two focal chords at right angles to each other is constant. or 194. The equation of the ellipse, referred to the vertex, is Hence, in the ellipse, the square of the ordinate is less than the rectangle under the parameter and abscissa. The equation of the byperbola is found in like manner, |