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2023年高考数学上海春20

  2023-07-08 14:37:36  

(18分)已知椭圆Γ:x2m2+y23=1(m>0m3)
(1)若m=2,求椭圆Γ的离心率;
(2)设A1A2为椭圆Γ的左右顶点,椭圆Γ上一点E的纵坐标为1,且EA1EA2=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y25m2x25=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abc,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=23m2p2m2+1=2,求解即可;
(3)设直线y=3x+t,与椭圆方程联立可得t23m2+3,与双曲线方程联立可得t2=5m215,可求m的取值范围.
解:(1)若m=2,则a2=4b2=3c=\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△=0t^{2}=5m^{2}-15
\therefore 5m^{2}-15\leqslant 3m^{2}+3
\therefore -3\leqslant m\leqslant 3
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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