优优班--学霸训练营 > 知识点挑题
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            • 1.
              已知线段\(AB\)垂直于定圆所在的平面,\(B\),\(C\)是圆上的两点,\(H\)是点\(B\)在\(AC\)上的射影,当\(C\)运动,点\(H\)运动的轨迹\((\)  \()\)
              A.是圆
              B.是椭圆
              C.是抛物线
              D.不是平面图形
            • 2.
              棱长为\(1\)的正方体\(ABCD-A_{1}B_{1}C_{1}D_{1}\),动点\(P\)在其表面上运动,且与点\(A\)的距离是\( \dfrac {2 \sqrt {3}}{3}\),点\(P\)的集合是一条曲线,则这条曲线的长度是\((\)  \()\)
              A.\( \dfrac {2 \sqrt {3}}{3}π\)
              B.\( \dfrac {5 \sqrt {3}}{6}π\)
              C.\( \sqrt {3}π\)
              D.\( \dfrac {7 \sqrt {3}}{6}π\)
            • 3.
              一动圆\(P\)过定点\(M(-3,0)\),且与已知圆\(N\):\((x-3)^{2}+y^{2}=16\)外切,则动圆圆心\(P\)的轨迹方程是\((\)  \()\)
              A.\( \dfrac {x^{2}}{4}- \dfrac {y^{2}}{5}=1(x\geqslant 2)\)
              B.\( \dfrac {x^{2}}{4}+ \dfrac {y^{2}}{5}=1(x\geqslant 2)\)
              C.\( \dfrac {x^{2}}{4}- \dfrac {y^{2}}{5}=1(x\leqslant -2)\)
              D.\( \dfrac {x^{2}}{4}+ \dfrac {y^{2}}{5}=1(x\leqslant -2)\)
            • 4.
              如图,各棱长均为\(1\)的正三棱柱\(ABC-A_{1}B_{1}C_{1}\),\(M\)、\(N\)分别为线段\(A_{1}\)B、\(B_{1}C\)上的动点,若点\(M\),\(N\)所在直线与平面\(ACC_{1}A_{1}\)不相交,点\(Q\)为\(MN\)中点,则\(Q\)点的轨迹的长度是\((\)  \()\)
              A.\( \dfrac { \sqrt {2}}{2}\)
              B.\( \dfrac { \sqrt {3}}{2}\)
              C.\(1\)
              D.\( \sqrt {2}\)
            • 5.
              阿波罗尼斯\((\)约公元前\(262-190\)年\()\)证明过这样一个命题:平面内到两定点距离之比为常数\(k(k > 0\)且\(k\neq 1)\)的点的轨迹是圆\(.\)后人将这个圆称为阿氏圆\(.\)若平面内两定点\(A\),\(B\)间的距离为\(2\),动点\(P\)与\(A\),\(B\)距离之比为\( \sqrt {2}\),当\(P\),\(A\),\(B\)不共线时,\(\triangle PAB\)面积的最大值是\((\)  \()\)
              A.\(2 \sqrt {2}\)
              B.\( \sqrt {2}\)
              C.\( \dfrac {2 \sqrt {2}}{3}\)
              D.\( \dfrac { \sqrt {2}}{3}\)
            • 6.
              当点\(P\)在圆\(x^{2}+y^{2}=1\)上变动时,它与定点\(Q(3,0)\)相连,线段\(PQ\)的中点\(M\)的轨迹方程是\((\)  \()\)
              A.\((x-3)^{2}+y^{2}=1\)
              B.\((2x-3)^{2}+4y^{2}=1\)
              C.\((x+3)^{2}+y^{2}=4\)
              D.\((2x+3)^{2}+4y^{2}=4\)
            • 7.
              若长度为定值\(4\)的线段\(AB\)的两端点分别在\(x\)轴正半轴和\(y\)轴正半轴上移动,\(P(x,y)\)为\(\triangle OAB\)的外心轨迹上一点,则\(x+y\)的最大值为\((\)  \()\)
              A.\(1\)
              B.\(4\)
              C.\( \sqrt {2}\)
              D.\(2 \sqrt {2}\)
            • 8.
              已知两定点\(F_{1}(0,-5)\),\(F_{2}(0,5)\),平面内动点 \(P\)到\(F_{1}\)、\(F_{2}\)的距离之差的绝对值是\(6\),则点\(P\)的轨迹方程为\((\)  \()\)
              A.\( \dfrac {x^{2}}{9}- \dfrac {y^{2}}{16}=1\)
              B.\( \dfrac {x^{2}}{16}- \dfrac {y^{2}}{9}=1\)
              C.\( \dfrac {y^{2}}{9}- \dfrac {x^{2}}{16}=1\)
              D.\( \dfrac {y^{2}}{16}- \dfrac {x^{2}}{9}=1\)
            • 9.
              方程\((x^{2}+y^{2}-4) \sqrt {x+y+1}=0\)的曲线形状是\((\)  \()\)
              A.
              B.
              C.
              D.
            • 10.
              已知分别过点 \(A(-1,0)\)和点 \(B(1,0)\)的两条直线相交于点\(P\),若 两 直 线 的 斜 率 之 积 为\(-1\),则 动 点\(P\)的 轨 迹 方 程 是\((\)  \()\)
              A.\(x^{2}+y^{2}=1\)
              B.\(x^{2}+y^{2}=1(x\neq ±1)\)
              C.\(x^{2}+y^{2}=1(\) \(x\neq 0)\)
              D.\(y= \sqrt {1-x^{2}}\)
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