高中数学-高考试题(上海交通大学 1931年旋转曲面方程)

问答题（1931年上海交通大学）

Find the equation of the sphere which passes through (1,5,-3),(-3,0,0) which center on the 3x+y+z=0,x+2y +1 = 0.

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讨论

旋转曲面方程

在xOy平面上，四边形ABCD的四个顶点坐标依次为(0,0),(1,0),(2,1)及(0,3).求这个四边形绕x轴旋转一周所得到的几何体的体积.

已知 ABCD 是边长为 1 的正方形, 绕其中一条轴 AB 旋转成一个圆柱.(1) 求该圆柱的表面积;(2) 将 DC 旋转 90° 至 C1D1, 求线 C1D 与平面 ABCD 的夹角.

Find the sum of n terms of the series whose nth term is 3(4n+4n²)-5n³.

Find the general term and the sum ofn terms of the series -3,-1,11,39,89,167.

Six persons throw for a stake, which is to be won by the one who first throws head with a coin: if they throw in succession, find the chance of the fourth person.

Find the maximum value of (5+x)(2+x)/(1-x).

Find all the positive angles less than 380° which satisfy the equation sinx+sin2xsin3x = 0.

Find the maximum value of (7-x)4 (2+x)6 when x lies between 7 and 2.

A man borrows $5000 at 4 percent compound interest, if the principal and interest are to be repaid by 10 equal instalments, find the amount of each instalment; having given 1g 1.04 =0.0170333 and lg675565=5.829667.

A,B,C are the angles of a triangle, prove that tanA+tanB+tanC=tanAtanBtanC.

曲面方程

如图，OA是圆锥底面中心O到母线的垂线,OA绕轴旋转一周所得曲面将圆锥分成体积相等的两部分，则母线与轴的夹角为【】

如图所示，为了制作一个圆柱形灯笼，先要制作4 个全等的矩形骨架，总计耗用9.6 米铁丝。骨架将到柱底面8 等分，再用S 平方米塑輯片制成圆柱的侧面和下底面（不安装上底面).（Ⅰ）当圆柱底面半径r 为何值时， S 取得最大值？并求出该最大值（结果精确到0.01 平方米）；（Ⅱ）在灯笼内，以矩形骨架的頂点为端点，安装一些霓虹灯，当灯笼底面半径为0.3 米时，求图中两根直线型霓虹灯A1B3,A3B5所在异面直线所成角的大小（结果用反三角函数值表示).

Find the equation of the ruled surface whose generators are the system of the lines x-2y =4kx,k(x-2y)=4 and discuss the surface.

Solve the following equation by taking the steps performed in srriving at Cardann formulas, but do not formulas themselves: x³ - 15x² - 33x +847= 0.

A boy standing cft. behind and opposite the middle of a foothall goal sees that the angle of elevation of the nearer is A and the angle of elevation of the farther one is B. Show that the length of the field is c(tanAcotB-1).

Find the equation of the projection of the linex=z+2,y=2z-4 upon the plane x+y- z = 0.

Given the parabola y²=4x and the line x=2+ecosα,y=-4+ecosβ，find the condition which cosα and cosβ must satisfy if the line meets the parabola in but one point.

Find the equation of the plane passing through the line (x-x1)/a1 =(y-y1)/b1 =(z-z1)/c1 which is parallel to the (x-x2)/a2 =(y-y2)/b2 =(z-z2)/c2 .

Chords of the circle p=29cosθ which pass through the pole are extended a distance 2b. Find the locus of their extremities.

Find the sum of the geometical series -2,,-1/3 to 6 terms.

空间解析几何

如图，在三棱锥S-ABC中，S在底面上的射影N位于底面的高CD上；M是侧棱SC上的一点，使截面MAB与底面所成的角等于∠NSC，求证：SC垂直于截面MAB.

如图，设平面AC和BD相交于BC，它们所成的一个二面角为45°，P为面AC内的一点，Q为面BD内的一点.已知直线MQ是直线PQ在平面BD内的射影，并且M在BC上，又设PQ与平面BD所成的角为β，∠CMQ=θ(0°<θ<90°)，线段PM的长为a.求线段PQ的长.

如图，已知二面角α-AB-β的平面角是锐角，C是平面α内的一点（它不在棱AB上），点D是点C在平面β上的射影，点E是棱AB上满足∠CEB为锐角的任意一点，那么【】.

如图，四棱锥S-ABCD的底面是边长为1的正方形，侧棱SB垂直于底面，并且SB=，用α表示∠ASD，求sinα的值.

如图，在三棱锥S-ABC中，SA⊥底面ABC,AB⊥BC.DE垂直平分SC，且分别交AC,SC于D,E.又SA=AB,SB=BC.求以BD为棱,以BDE与BDC为面的二面角的度数.

如图，平面α,β相交于直线MN，点A在平面α上，点B在平面β上，点C在直线MN上，∠ACM=∠BCN=45°，A-MN-B是60°的二面角，AC=1. 求：(1) 点A到平面β的距离；(2) 二面角A-BC-M的大小（用反三角函数表示）.

如图，已知ABCD是边长为4的正方形，E,F分别是AB,AD的中点，GC垂直于ABCD所在的平面，且GC=2.求点B到平面EFG的距离.

如图，在正三角棱柱ABC-A1 B1 C1中，E∈BB1，截面A1 EC⊥侧面AC1 （Ⅰ）求证: BE=EB1;（Ⅱ）若AA1=A1 B1，求平面A1 EC与平面A1 B1 C1所成二面角（锐角）的度数. 注意：在下面横线上填写适当内容,使之成为(Ⅰ)的完整证明,并解答(Ⅱ).（Ⅰ）证明:（如图）在截面A1 EC内，过E作EG⊥A1 C,G是垂足. ①∵_________________________________________∴EG⊥侧面AC1；取AC的中点F，连接BF,FG,由AB=BC得BF⊥AC.②∵_________________________________________.∴BF⊥侧面AC1；得BF//EG,BF、EG确定一个平面，交侧面AC1于FG.③∵_________________________________________∴BF//EG四边形BEGF是平行四边形BF=EG.④∵_________________________________________∴FG//AA1,ΔAA1 C∽ΔFGC.⑤∵_________________________________________∴FG=1/2 AA1=1/2 BB1,即BE=1/2 BB1故BE=EB1.（Ⅱ）解：

如图，已知正四棱锥ABCD-A1 B1 C1 D1，点E在棱D1 D上，截面EAC//D1 B，且EAC与底面ABCD所成角为45°,AB=a. (Ⅰ)求截面EAC的面积；(Ⅱ)求异面直线A1 B1与AC之间的距离；(Ⅲ)求三棱锥B1-EAC的体积.

如图，已知平行六面体ABCD-A1B1C1D1的底面ABCD是菱形，且∠C1CB=∠C1CD=∠BCD=60°. (Ⅰ)证明：C1C⊥BD.(Ⅱ)假设CD=2,CC1=3/2,记面C1BD为α，面CBD为β，求二面角a-BD-β的平面角的余弦值.(Ⅲ)当CD/CC1 的值为多少时，能使A1C⊥平面C1BD？请给出证明.