2 7. Find the sum of either of the two following series: (1) sin + sin 20+ sin 30+ &c....+ sin no. (2) cos 0+ cos 20+ cos 30+ &c....+ cos no. Tod. Trig., Arts. 303, 304. 8. If A be the vertex of a parabola, S the focus, Pany point on the curve, T the point where the tangent at P meets the axis produced, PQ perpendicular to the directrix, prove (1) PT bisects the angle SPQ, (2) the quadrilateral SPQT is a rhombus. Drew, Geom. Conics, Props. v., VIII. 9. Prove that the perpendiculars drawn from the foci of an ellipse on a tangent at any point intersect the tangent in the circumference of the circle described on the major axis. Drew, Conic Sections, Prop. xv. If a tangent be drawn at the extremity of a latus rectum of the ellipse, find the lengths of the perpendiculars upon this tangent from the foci. By similar triangles (fig. 10) we have SY: HZ:: SL: HL; 10. If a right cone be cut by a plane which meets the cone on both sides the vertex, the section is an hyperbola. Drew, Conic Sections, Props. of Cone. 11. Given the equations to two straight lines referred to rectangular coordinates, find the angle at which they intersect. Salmon, Conic Sections, Art. 25. Prove that the equation 3y2 - 8xy-3x2 - 29x + 3y - 18=0 represent two straight lines at right angles to each other. This equation is equivalent to (3y+x+9) (y - 3x-2) = 0, which factors represent two straight lines at right angles to each other. represents an ellipse. Find its axes and determine its area. Its centre is at the point (1, 1). Transforming equation to parallel axes through this point, we have the curve represented by 5x+5y+2xy-12=0. Now, to determine the directions of the principal axes, we have tan 20 = ∞ 0 = 45°. Now, transforming the equation to axes through same origin as last, but making an angle of 45° with them, our equation takes the form 3x+2y=6 (see fig. 11); therefore the semi-axes are √3 and √2, and the area =√/6π. 13. When is an hyperbola said to be equilateral or rectangular? Find the eccentricity of such an hyperbola. In a rectangular hyperbola the distance of any point of the curve from the centre is a mean proportional between the distances of the same point from the foci. Salmon, Conic Sections, Art. 174. If in any hyperbola CD be conjugate to CP, we have SP.HP=CD, but in the rectangular hyperbola any diameter is equal to its conjugate; therefore CP-CD, and we have SP. HP=CP2. PURE MATHEMATICS (2). TUESDAY, 12TH DECEMBER, 1876. 10 A.M. TO 1 P.M. 1. Prove that the sum of the squares on the three . straight lines drawn from the angles of a triangle to the points of bisection of the opposite sides is equal to threefourths of the sum of the squares of the sides of the triangle. Let ABC be the triangle; D, E, F the middle points of the sides. Then we have similarly and AB2 + BC2 = 2AE2 + 2EB2; AB2 + AC2 = 2AD2 +2CD2; and, by addition, 2 (AB2 + BC2 + CA2) = · 2 (AD2 + BE2 + CF2) + 1⁄2 (AB2 + BC2+CA2); therefore AD2 + BE2 + CF2 = 3 (AB2 + BC2 + CA2). 2. Given one side of a triangle, the angle opposite to that side and the point at which the inscribed ́circle touches that side, construct the triangle. E Let AB (fig. 12) be the given side of the triangle. On AB describe a segment of a circle containing an angle equal to the given angle; and also describe a segment ADB containing an angle equal to half the given angle and a right angle. With centre A and distance equal to the difference between the segments of the line given, describe a circle cutting latter segment in D. Join AD, and produce it to meet the outer segment in C. Join CB. Then ACB is the triangle required. Join DB. We shall find CD = CB, and therefore difference between AC and CB= difference of segments of base, and angle ACB= vertical angle required; therefore ACB is the triangle required. 3. Semicircles BDA and CEA are described upon the sides BA and AC of the right-angled triangle BAC, and another semicircle BFAGC is described upon the hypothenuse BC; prove that the sum of the areas of the two curvilinear figures BDAF and CEAG is equal to the area of the triangle ABC. The whole semicircle BFAGC is equal to the sum of the two semicircles BDA and AEC; therefore, taking away the common parts AFB and AGC, we have the triangle ABC equal to the two curvilinear areas BDAF and CEAG. 4. If a=xy11, b=xy11, c=xy"`, prove that a2-br-3c2¬2 = (xyp-1)?" (xy11)"-"(xy"-1)-2 = x°y° = 1. If 2s=a+b+c, prove that ( s − a)3 + ( s − b) 3 + (s − c)3 + 3abc = s3. Put x=s-a, y=s—b, z=s—c, then x+y+z=s. Then, since x3 + y3 + z3 + 3 (y + z) (z + x) (x + y) = x3 + y3 + z3 + 3 (yz2 + y2z + zx2 + z*x + xy2 + x2y) + 6xyz therefore (sa)3 + (s − b)3 + ( s − c)3 + 3abc=s3. 5. If x be a possible quantity, prove that must be either < 1 or > 3. 2x2+ bx+3 2x+1 Putting this expression =y and solving for x, we y-3 √(y-1) (y-3)}, which shews obtain x 2 = ± 2 that the values of x are not possible if y> 1 and <3, and that therefore for all real values of x, y has no value between these limits. th 6. If the m term of a harmonic progression be equal to n, and the nth term be equal to m, prove that the (m + n)th term is equal to nm m+ n The mth and nth terms of the corresponding arith 1 1 metic series will be and and therefore the (m +n)th n m term of the arithmetic series will be = m + n and there mn fore the (m+n)th term of the Harmonic Series being |