2023年高考数学上海春20 |
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2023-07-08 14:37:36 |
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(18分)已知椭圆Γ:x2m2+y23=1(m>0且m≠√3). (1)若m=2,求椭圆Γ的离心率; (2)设A1、A2为椭圆Γ的左右顶点,椭圆Γ上一点E的纵坐标为1,且→EA1⋅→EA2=−2,求实数m的值; (3)过椭圆Γ上一点P作斜率为√3的直线l,若直线l与双曲线y25m2−x25=1有且仅有一个公共点,求实数m的取值范围. 分析:(1)由题意可得a,b,c,可求离心率; (2)由已知得A1(−m,0),A2(m,0),设E(p,1),由已知可得p2=23m2,p2−m2+1=−2,求解即可; (3)设直线y=√3x+t,与椭圆方程联立可得t2⩽3m2+3,与双曲线方程联立可得t2=5m2−15,可求m的取值范围. 解:(1)若m=2,则a2=4,b2=3,∴,c=\sqrt{{a}^{2}-{b}^{2}}=1,\therefore e=\dfrac{c}{a}=\dfrac{1}{2}; (2)由已知得A_{1}(-m,0),A_{2}(m,0),设E(p,1), \therefore\dfrac{{p}^{2}}{{m}^{2}}+\dfrac{1}{3}=1,即p^{2}=\dfrac{2}{3}m^{2}, \therefore\overrightarrow{E{A}_{1}}=(-m-p,-1),\overrightarrow{E{A}_{2}}=(m-p,-1),\therefore\overrightarrow{E{A}_{1}}\cdot \overrightarrow{E{A}_{2}}=(-m-p,-1)\cdot (m-p,-1)=p^{2}-m^{2}+1=-2, \because p^{2}=\dfrac{2}{3}m^{2},代入求得m=3; (3)设直线y=\sqrt{3}x+t,联立椭圆可得\dfrac{{x}^{2}}{{m}^{2}}+\dfrac{(\sqrt{3}x+t)^{2}}{3}=1, 整理得(3+3m^{2})x^{2}+2\sqrt{3}tm^{2}x+(t^{2}-3)m^{2}=0, 由△\geqslant 0,\therefore t^{2}\leqslant 3m^{2}+3, 联立双曲线可得\dfrac{(\sqrt{3}x+t)^{2}}{5{m}^{2}}-\dfrac{{x}^{2}}{5}=1,整理得(3-m^{2})x^{2}+2\sqrt{3}tx+(t^{2}-5m^{2})=0, 由△=0,t^{2}=5m^{2}-15, \therefore 5m^{2}-15\leqslant 3m^{2}+3, \therefore -3\leqslant m\leqslant 3, 又5m^{2}-15\geqslant 0,\therefore m\geqslant \sqrt{3},\because m\ne \sqrt{3}, 综上所述:m\in (\sqrt{3},3]. 点评:本题考查离心率的求法,考查椭圆与双曲线的几何性质,直线与椭圆的综合,属中档题.
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