x 1 F 31 – 2x, which, after proper reductions, will give 2 (x2 + yo - rx) = p2 { (ic - ) = yo}. r = 104. Let x = the part prochuced of the given line a ; then by the question (a + x). a = a + ax ao ata 5 2 a = (1 + 75) which is the production required. it. 105. Let AB (Fig. 58,) = a, be the given line, required to construct a Tetraedron upon In the plane ABC construct the equilateral A ABC, and let O be the point in it equally distant from A, B, C, and make OD I plane ABC = a ; then AD, BD, CD, being joined, the ✓ 3 figure ABCD will be the Tetraedron required. For AO : BA :: sin. 30° : sin, 60° N 3 :: 1 : 73 2 + Hence AD = AO? + DO? = 2a2 = a, and simi 3 larly DC, BD, may be shewn to be = a. :. the four faces are equal equilateral A &c. Again, r being taken in OD so that rA = 1D, we have AO + Or? = Ar = „D” = AO? + (OD – D = AO” + OD + D - 2 OD X D. :. 2 OD X rD = AO + OD = AD = a* or 7D = a 2 which is the radius of the circum20D 2a } scribius sphere. 106. Tu and the focus of a given parabola, either use the method adopted ia Prob. 95, or the following one ; drawing two perallel lines in the parabola QV, Q'V', and let them be bisected in V, V', join VV' and produce it to meet the curve in P, thus determining an abscissa PV, corresponding to the ordinate QV. In like manner drawing any two parallel lines not parallel to the former, we get another pair of oblique co-ordinates pv, qu. Now by property of the curve, we have (S being the focus) QV2 .. SP QV? = 45P X PV) 4PV being known. if and qué = 4 Sp x pv 4pv with P, p, as centres, and radii = SP, Sp, we describe two circular arcs, their intersection will be the focus required. qu? p = 107. From similar A OPM, ONQ, and the property of the parabola, we have AN : AM :: QN? : PM? :: ON : OM? :: (AO — AN) : (AM – AO)? QN: :. AN X (AM – 240 x AM + A0%) = AM ~ (AN? 2A0 X AN + A0%), and by reduction AO® X (AM – AN) = AN. AM (AM – AN) 108. Let AP (Fig. 47,) be any conic section whose focus is S, and vertex A; also let La, Aa, be the tangents at the extremity of the latus rectum SL, and vertex A intersecting in a; then Aa = AS. 6 For supposing y = ✓ 2ax F xito be the equation to the conic a a section, (which will be an ellipse, parabola, or circle, when the negative sign is taken, and an hyperbola when the positive,) a and b being axes, and x, y, originating in A, we have. y' – y= dy ( (a' - x). (1) the equation to the tangent dx b dy at any point. Hence, since a F x dx 2ax F x2 6 ý = a F x bx x + ✓ 2ax F x2 2ax Fx® Let x = AS = m, and a' = 0; Then Aa = y = : ; a v 2am Fm But a’ = b + (a F m) = b + a® + m2 F 2am from wellknown property of the curve, i. 2am F mi = 62 a b.m 109. If R, r, be the radii of the circles, circumscribing and inscribing a A whose sides are a, b, c, then abc Rr = For, supposing A, B, C, the angles to which a, b, c, are respectively opposite, since those angles are each bisected by the lines joining them and the centre of the inscribed 0, we have B с a=pcot. + r cot. 2 a + 6 a + b + c A с 2 2 a + b cot. 422 2 2 2ab al - 62 + ca 4ab which gives by reduction 4r2 = {c2 – (a - b)2} l. a + b + c Again, since the chord of an arc = 2 sin. of half that arc, we have 2C 2 c? (2) (a+b+c).(a+b-c) (c+6-a). (c+a-6) a b c :. 4R' go= (a + b + c) abc and Rr = Q.E.D. 2 (a+b+c) 110. The equation to an equilateral hyperbola whose axes are unity, and co-ordinates at the centre being y = V1 + x2 2 2 2 we have s ydx = sdx 1 + x - l. (x + N 1 + x?) :: by the question :: x + 1+x2 = (x' + N 1 + x'?)" N 1 + x2 x + 1 + x (x + 1+ 2?)? 1 (x' + 1+x2) (x' + N 1+x)? 1 2 1 .. = (1=); 212 111. Let LP (Fig. 47,) be a tangent at L the extremity of the latus rectum of the ellipse, meeting any ordinate PN produced in P'; then S being the focus, required to shew that SP = NP'. a and b being the semi-axes of the ellipse, y, x, its co-ordinates at the vertex, and y', ', those of the tangent at any point, originating also at the vertex, we have b 2ax x2 a and y' bx b x x + V 2ax Eva' - 63 x' + a + va — b? = |