Let CF, or CF produced, meet AB in G. Draw GH parallel to EC or DF. Then, by (4), GH is at right angles to the plane CAB; and therefore / AGH is a right angle. Similarly, a line GH' drawn parallel to ED will lie in the plane of ECD, and will be at right angles to the plane DAB. Therefore ▲ AGH' is a right angle. Hence AB will be at right angles to the plane in which GH, GH' lie, (XI. 4), and therefore at right angles with the line CFG, which also lies in that plane. 1849. (4). Draw a straight line perpendicular to a plane from a given point without it. (XI. 11.) (B). Prove that equal right lines drawn from a given point to a given plane are equally inclined to the plane. Let P be the given point; PA, PA', two equal straight lines drawn from P to meet the plane. As in (A) draw a perpendicular PN to the plane. Join NA, NA'. By the definition of a perpendicular to a plane, PN makes right angles with every line in the plane drawn from N, and therefore with NA, NA'. Hence in the right-angled triangles PAN, PA'N, PA = PA', therefore and PN is common ; NA NA', (1. 47), PAN PA'N, and the triangles are equal. Hence = or the lines PA, PA' are equally inclined to the plane. GEOMETRICAL CONIC SECTIONS. 1848. (4). Assuming the tangent at any point P of a parabola to make equal angles with the focal distance SP and the diameter at that point, prove that SY, the perpendicular upon it from the focus, meets it in the tangent at the vertex. (B). If PM be the ordinate at P, and T the intersection of the tangent at P with the axis, TP.TY TM.TS. Since, by (4), SYT (fig. 15) is a right angle, the triangles STY, PTM are similar. Hence therefore Or thus: TY: TS:: TM: TP, TP.TY= TM.TS. Since SYP, SMP are right angles, a circle can be described about SYPM. Therefore 1851. TP.TY = TM.TS. (4). Assuming that the sum of the focal distances of a point in the ellipse is equal to a given line, shew that the axis-major is equal to the same line. (B). Shew that the axis-major is greater than any other diameter. Let PCP' (fig. 16) be any diameter. Join P and P' with the foci. and Then, since two sides of a triangle are greater than the third, SP + SP' > PP', HP + HP' > PP' ; .. SP + HP + SP' + HP' > 2PP'. 1848. (4). If one of the focal distances SP of a point P be produced to L, a straight line PT which bisects the exterior angle HPL is the tangent to the curve at P. (B). For what position of P is the angle SPH greatest? By (A), (fig. 17), ▲ SPH + 24 SPT = 2 right angles. Therefore SPH is greatest when SPT is least. Now as P moves from A to B, 4SPT decreases from a right angle to SBt (where Bt || CT); and as P moves from B to A', ▲ SPT increases from SBt to a right angle. Thus SPT is least, and therefore SPH is greatest, when P is at B. 1851. (A). In the ellipse if PU be a tangent at P, meeting the minor axis produced in U, and PN be drawn perpendicular to the minor axis, then CN: CB:: CB: CU. (B). If a series of ellipses be described having the same major axis, the tangents at the extremities of their laterarecta will all meet the minor axis in the same point. Let SL (fig. 18) be the semi-latus-rectum of one of the ellipses, U the point where the tangent at L meets the minor axis. By (4), But therefore and therefore CU: CB:: CB : SL. CB: SL:: CA: CB; CU: CB:: CA: CB, CU = CA, which is constant, since all the ellipses have the same major axis. 1850. 1849. (4). The perpendiculars from the foci on the tangent of an ellipse intersect the tangent in the circumference of a circle having the axis-major as diameter. (B.) Employ this proposition to find the locus of the intersection of a pair of tangents at right angles to each other. (C). Deduce from the proposition an analogous one for the parabola. (B). Let SY, HZ (fig. 19) be the perpendiculars from the foci on the tangents; Y, Z being by (4) points in circumference of the circle whose diameter is AA'. If we produce YC to Z, and join HZ, we obtain from the triangles CSY, CHZ, that HZ, = SY, that HZ, is in the same straight line with ZH, and therefore that SY.HZ = ZH.HZ = A'H.HA = BC2. Let the tangent at P meet the tangent which is at right angles to it in Q, which latter tangent suppose intersects the circle in Y, Z. Join SY', HZ. By (A), SY', HZ' are perpendicular to QZ'; and therefore QY' = SY, QZ' = HZ. Hence, drawing QKK' through the centre C, QK.QK' = QY'.QZ' SY.HZ BC2. Therefore the distance of Q from C is constant, and the locus of Q is a circle whose centre is C. (C). The ellipse will become a parabola if C moves off to an infinite distance, while the vertex A' and the focus H remain fixed. For, if C and therefore S move off to an infinite distance, PS will become parallel to A'A. Hence the tangent ZPY, which always makes equal angles with HP and PS, will in this case make equal angles with HP and a line through P parallel to A'A; and the curve will therefore be a parabola. Now the locus of Z in the ellipse is a circle whose diameter is A'A; and when C moves off to an infinite distance, the radius of this circle becomes infinite, and the circle coincides with its tangent at the point A'. Consequently, in the parabola, the locus of the intersection of the tangent with the perpendicular upon it from the focus is the tangent at the vertex. 1848. (A). In an ellipse the sum of the squares of any two conjugate diameters is invariable. (B). When is the square of their sum least? 2 (CP+ CD) = CP* + CD +2CP.CD, (fig. 20) = AC2 + BC2 + 2CP.CD, by (4). Therefore (CP+ CD)2 will be least when CP.CD is least. CP. CP PF' therefore CP.CD is least when is least; i.e. when PF = CP, or when CP, CD are at right angles. Hence (CP + CD)2 is least when CP, CD coincide with CA, CB. 1848. (A). Prove that all parallelograms whose sides touch an ellipse at the ends of conjugate diameters are equal. (B). Prove that such parallelograms have the least area of all which circumscribe the ellipse. Let TTTT (fig. 21) be a parallelogram circumscribing an ellipse at the extremities of conjugate diameters CP, CD; t ̧tëtât another parallelogram circumscribing the ellipse. Draw PF perpendicular to DCD'. Then area of TTTT = 2PF.DD', area of tttt 2PF.dd'. 1234 1 2 3 4 Now DD' is the least possible value of dd'; therefore area of tttt is always greater than that of T,T,T,T, 1 2 4. |