97. The semi-axes are the positive values of r found from direction cosines are l, m, n, the area of the section is known to be πατε √(l3a2 + m2b2 +n3c2)* Now if we seek the maximum and minimum values of l'a2 + m2b2 + n2c2, with the condition that the plane is to thus the product of the maximum and minimum values is 103. The process will resemble that given in Gregory's Solid Geometry for finding the area of the section of the ellipsoid by the plane lx + my + nz = 0. If a, ß, y be co-ordinates of the centre of (a3l3 + b3m2 + c3n2 – 82) πabc See also Differential Calculus, Chapter XVI. 104. The first part may be proved by taking the general equa tion ax2 + by3 + cz2 + 2a'yz + 26'zx + 2c'xy + 2a′′x + 26′′y + 2c′′z +ƒ = 0; and the cutting planes may be supposed parallel to that of (x, y). For the latter part we must find the ratio of similarly situated lines in the two sections, and the square of this ratio will be the ratio of the areas. 105. This is a mode of expressing the result of the two preceding examples. The result may also be obtained very easily by referring the ellipsoid to conjugate diameters, two of which are in a plane parallel to the planes considered. 106. The equation to the ellipsoid is ха y3 3 Ха + a2 + + = + where (a, B, y) is the external point. 109. Hyperboloid of one sheet. one sheet. 110. Hyperboloid of 111. Hyperbolic paraboloid. 113. Hyperboloid of two sheets 112. Parabolic cylinder. if ƒ be positive. 114. Right circular cylinder if ƒ be positive. 115. Hyperboloid of revolution of two sheets if ƒ be greater than 48, cone if ƒ = 48, hyperboloid of revolution of one sheet ifƒ be less than 48. 116. Hyperboloid of revolution of two sheets. 118. There is a line of centres, given by the equations k = 0, h−l+1 = 0. 122. A surface of revolution, the axis of which is x = y = 2. 124. Use the equation in Example 65; the locus is a sphere, the radius of which is (a+b+c2 + h2 + k3). 133. The sine of the angle is scending order of magnitude. in Geometry of Two Dimensions, Plane Co-ordinate Geometry, Art. 286. 144. Refer the ellipsoid to conjugate diameters such that the plane containing two of them is parallel to the cutting plane. 157. b2 - ac must be positive and (b’2 — ac) (a’2 — bc)=(a'b′—cc')3. 159. Let x cos 0, + y cos 0 ̧ + z cos 01⁄2 1 2 = Ղ, =v; then the second equation is uv = 0, and the first is (u+v) (u−v)=0. The line of intersection will be found to be determined by 161. 2 = 0; y = 0, zc - xa = 0 ; The cosine of the inclination of the plane to the axis. 170. Take the equation to an enveloping cylinder and apply the tests in order that it may be a surface of revolution. 182. (x, y, z) = ☀ (a + x − a, ß + y − ẞ, y + z − y) The equation to the tangent plane to the surface (x, y, z) = 0 at the point (x, y, z) is do 0; dz = d'u d3u +(x − a) (y-ẞ) Now suppose this plane to pass through the point (a, B, y), and we obtain a relation which by means of the equation 4 (x, y, z) = 0, transformed as above, reduces to 183. Let a line pass through the point (a, ẞ, y) and through the point (x, y, z) on the surface; and let (x, y, z) be any other point on this line; then we have Substitute these values of x, y, z in 4 (x, y, z) = 0; we thus obtain a quadratic in r corresponding to the two points in which the line through (a, ẞ, y) and (x', y', x') meets the surface. The condition that this quadratic should have equal roots leads to the equation to the required cone, x, y, z being the variable co-ordinates. 205. Take for the equations to the lines those in example 74; let a be the radius of the circle; the required equation is (mxc3 — cyz)2 + m2 (mcxz — c3y)2 = m2a2 (c2 — z3)2. 212. (1) is not a developable surface; (2) is. 218. (ax + by + cz)3 = r2 {(x + a)2 + (y + b)2 + (z+c)3}. 227. Use conjugate diameters as axes, two of them being parallel to the plane of the circles, and the third passing through the centres of the circular sections. 233. The curve is determined by the given equation combined through the line of intersection of the normal plane at the point (x, y, z) and the consecutive normal plane, and perpendicular to the first normal plane. 243. (h3a2 + k2x2) + 246. A circle. |