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2025年高考数学北京-18<-->2025年高考数学北京-20
（15分）已知椭圆$E:\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1(a > b > 0)$的离心率为$\dfrac{\sqrt{2}}{2}$，椭圆$E$上的点到两个焦点的距离之和为4．
（1）求椭圆$E$的方程；
（2）设$O$为原点，$M(x_{0}$，$y_{0})(x_{0}\ne 0)$在椭圆$E$上，直线$x_{0}x+2y_{0}y-4=0$与$y=2$和$y=-2$分别交于$A$，$B$两点．设△$OAM$和△$OBM$的面积分别为$S_{1}$和$S_{2}$，比较$\dfrac{{S}_{1}}{{S}_{2}}$与$\dfrac{\vert OA\vert }{\vert OB\vert }$的大小．
〖答案〗（1）$\dfrac{x^{2}}{4}+\dfrac{y^{2}}{2}=1$；
（2）$\dfrac{S_{1}}{S_{2}}=\dfrac{\vert OA\vert }{\vert OB\vert }$．
〖分析〗（1）由椭圆定义得$a=2$，根据离心率得$c=\sqrt{2}$，则$b^{2}=2$，即可得椭圆方程；
（2）联立$\left\{\begin{array}{l}x_{0}x+2y_{0}y-4=0\\ \dfrac{x^{2}}{4}+\dfrac{y^{2}}{2}=1\end{array}\right.$，结合$\dfrac{x_{0}^{2}}{4}+\dfrac{y_{0}^{2}}{2}=1$，可得$y=y_{0}$，从而得直线$x_{0}x+2y_{0}y-4=0$与椭圆相切，$M$为切点，设$A(x_{1}$，$y_{1})$，$B(x_{2}$，$y_{2})$，当$x_{1}=x_{2}$时，易得$\dfrac{S_{1}}{S_{2}}=\dfrac{\vert OA\vert }{\vert OB\vert }$，进而设$x_{2} < x_{0} < x_{1}$，经运算可得$\dfrac{S_{1}}{S_{2}}=\dfrac{\vert OA\vert }{\vert OB\vert }$．
〖解答〗解：（1）由椭圆定义可知，$2a=4$，所以$a=2$，
又$e=\dfrac{c}{a}=\dfrac{\sqrt{2}}{2}$，所以$c=\sqrt{2}$，则$b^{2}=a^{2}-c^{2}=2$，
故椭圆方程为$\dfrac{x^{2}}{4}+\dfrac{y^{2}}{2}=1$；
（2）由题意，可得$\dfrac{x_{0}^{2}}{4}+\dfrac{y_{0}^{2}}{2}=1$，即$16-4x_{0}^{2}=8y_{0}^{2}$，
联立$\left\{\begin{array}{l}x_{0}x+2y_{0}y-4=0\\ \dfrac{x^{2}}{4}+\dfrac{y^{2}}{2}=1\end{array}\right.$，消去$x$，可得$(\dfrac{4-2y_{0}y}{x_{0}})^{2}+2y^{2}=4$，
整理得$(2x_{0}^{2}+4y_{0}^{2})y^{2}-16y_{0}y+16-4x_{0}^{2}=0$，
即$8y^{2}-16y_{0}y+8y_{0}^{2}=0$，则$(y-y_{0})^{2}=0$，所以$y=y_{0}$，
所以直线$x_{0}x+2y_{0}y-4=0$与椭圆相切，$M$为切点，
设$A(x_{1}$，$y_{1})$，$B(x_{2}$，$y_{2})$，
易知当$x_{1}=x_{2}$时，由对称性可知，$\dfrac{S_{1}}{S_{2}}=\dfrac{\vert OA\vert }{\vert OB\vert }$，
故设$x_{2} < x_{0} < x_{1}$，易知$\dfrac{S_{1}}{S_{2}}=\dfrac{\vert AM\vert }{\vert BM\vert }=\dfrac{\vert x_{1}-x_{0}\vert }{\vert x_{2}-x_{0}\vert }=\dfrac{x_{1}-x_{0}}{x_{0}-x_{2}}$，
联立$\left\{\begin{array}{l}x_{0}x+2y_{0}y-4=0\\ y=2\end{array}\right.$，解得$x_{1}=\dfrac{4-4y_{0}}{x_{0}},y_{1}=2$，
联立$\left\{\begin{array}{l}x_{0}x+2y_{0}y-4=0\\ y=-2\end{array}\right.$，解得$x_{2}=\dfrac{4+4y_{0}}{x_{0}},y_{2}=-2$，
所以$\dfrac{S_{1}}{S_{2}}=\dfrac{x_{1}-x_{0}}{x_{0}-x_{2}}=\dfrac{\dfrac{4-4y_{0}}{x_{0}}-x_{0}}{x_{0}-\dfrac{4+4y_{0}}{x_{0}}}=\dfrac{4-4y_{0}-x_{0}^{2}}{x_{0}^{2}-4y_{0}-4}=\dfrac{2y_{0}^{2}-4y_{0}}{-2y_{0}^{2}-4y_{0}}=\dfrac{2-y_{0}}{2+y_{0}}$，
$\dfrac{\vert OA\vert }{\vert OB\vert }=\dfrac{\sqrt{(\dfrac{4-4y_{0}}{x_{0}})^{2}+4}}{\sqrt{(\dfrac{4+4y_{0}}{x_{0}})^{2}+4}}=\dfrac{\sqrt{4(1-y_{0})^{2}+x_{0}^{2}}}{\sqrt{4(1+y_{0})^{2}+x_{0}^{2}}}=\dfrac{\sqrt{y_{0}^{2}-4y_{0}+4}}{\sqrt{y_{0}^{2}+4y_{0}+4}}=\dfrac{2-y_{0}}{2+y_{0}}$，
故$\dfrac{S_{1}}{S_{2}}=\dfrac{\vert OA\vert }{\vert OB\vert }$．
〖点评〗本题考查椭圆方程的求法，考查直线与椭圆的综合应用，属难题．
2025年高考数学北京-18<-->2025年高考数学北京-20
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