优优班--学霸训练营 > 知识点挑题
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            • 1.
              以椭圆短轴为直径的圆经过此椭圆的长轴的两个三等分点,则椭圆的离心率是\((\)  \()\)
              A.\( \dfrac {1}{3}\)
              B.\( \dfrac { \sqrt {3}}{3}\)
              C.\( \dfrac { \sqrt {3}}{4}\)
              D.\( \dfrac {2 \sqrt {2}}{3}\)
            • 2.
              已知点\(P(1,m)\)在椭圆\( \dfrac {x^{2}}{4}+y^{2}=1\)的外部,则直线\(y=2mx+ \sqrt {3}\)与圆\(x^{2}+y^{2}=1\)的位置关系为\((\)  \()\)
              A.相离
              B.相交
              C.相切
              D.相交或相切
            • 3.
              已知\(F\)是椭圆\(E: \dfrac {x^{2}}{a^{2}}+ \dfrac {y^{2}}{b^{2}}=1(a > b > 0)\)的左焦点,经过原点的直线\(l\)与椭圆\(E\)交于\(P\),\(Q\)两点,若\(|PF|=2|QF|\),且\(∠PFQ=120^{\circ}\),则椭圆\(E\)的离心率为\((\)  \()\)
              A.\( \dfrac {1}{3}\)
              B.\( \dfrac {1}{2}\)
              C.\( \dfrac { \sqrt {3}}{3}\)
              D.\( \dfrac { \sqrt {2}}{2}\)
            • 4.
              从集合\(\{2,4,8\}\)中随机选取一个数\(m\),则方程\( \dfrac {x^{2}}{m}+ \dfrac {y^{2}}{4}=1\)表示离心率为\( \dfrac { \sqrt {2}}{2}\)的椭圆的概率为\((\)  \()\)
              A.\( \dfrac {1}{4}\)
              B.\( \dfrac {1}{3}\)
              C.\( \dfrac {2}{3}\)
              D.\(1\)
            • 5.
              已知\(F_{1}(-1,0)\),\(F_{2}(1,0)\)是椭圆\(C\)的两个焦点,过\(F_{2}\)且垂直于\(x\)轴的直线交椭圆于\(A\)、\(B\)两点,且\(|AB|=3\),则\(C\)的方程为\((\)  \()\)
              A.\( \dfrac {x^{2}}{2}+y^{2}=1\)
              B.\( \dfrac {x^{2}}{3}+ \dfrac {y^{2}}{2}=1\)
              C.\( \dfrac {x^{2}}{4}+ \dfrac {y^{2}}{3}=1\)
              D.\( \dfrac {x^{2}}{5}+ \dfrac {y^{2}}{4}=1\)
            • 6. 椭圆C:+=1(a>b>0)的右顶点为A,P是椭圆C上一点,O为坐标原点.已知∠POA=60°,且OP⊥AP,则椭圆C的离心率为 ______
            • 7.
              设椭圆\( \dfrac {x^{2}}{m^{2}}+ \dfrac {y^{2}}{n^{2}}=1(m > 0,n > 0)\)的右焦点与抛物线\(y^{2}=8x\)的焦点相同,离心率为\( \dfrac {1}{2}\),则此椭圆的方程为\((\)  \()\)
              A.\( \dfrac {x^{2}}{12}+ \dfrac {y^{2}}{16}=1\)
              B.\( \dfrac {x^{2}}{16}+ \dfrac {y^{2}}{12}=1\)
              C.\( \dfrac {x^{2}}{48}+ \dfrac {y^{2}}{64}=1\)
              D.\( \dfrac {x^{2}}{64}+ \dfrac {y^{2}}{48}=1\)
            • 8.
              下列结论中,正确的有\((\)  \()\)
              \(①\)不存在实数\(k\),使得方程\(x\ln x- \dfrac {1}{2}x^{2}+k=0\)有两个不等实根;
              \(②\)已知\(\triangle ABC\)中,\(a\),\(b\),\(c\)分别为角\(A\),\(B\),\(C\)的对边,且\(a^{2}+b^{2}=2c^{2}\),则角\(C\)的最大值为\( \dfrac {π}{6}\);
              \(③\)函数\(y= \dfrac {1}{2}\ln \dfrac {1-\cos x}{1+\cos x}\)与\(y=\ln \tan \dfrac {x}{2}\)是同一函数;
              \(④\)在椭圆\( \dfrac {x^{2}}{a^{2}}+ \dfrac {y^{2}}{b^{2}}=1(a > b > 0)\),左右顶点分别为\(A\),\(B\),若\(P\)为椭圆上任意一点\((\)不同于\(A\),\(B)\),则直线\(PA\)与直线\(PB\)斜率之积为定值.
              A.\(①④\)
              B.\(①③\)
              C.\(①②\)
              D.\(②④\)
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