题目内容

5.如图,已知椭圆$E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1({a>b>0})$的离心率为$\frac{{\sqrt{2}}}{2}$,且过点$({2,\sqrt{2}})$,四边形ABCD的顶点在椭圆E上,且对角线AC,BD过原点O,${k_{AC}}•{k_{BD}}=-\frac{b^2}{a^2}$.
(1)求$\overrightarrow{OA}•\overrightarrow{OB}$的取值范围;
(2)求证:四边形ABCD的面积为定值.

分析 (1)由椭圆离心率为$\frac{{\sqrt{2}}}{2}$,且过点$({2,\sqrt{2}})$,列出方程组求出a,b,由此能求出椭圆的标准方程.
(2)设直线AC、BD的方程分别为$y=kx,y=-\frac{1}{2k}x$.联立$\left\{\begin{array}{l}y=kx\\ \frac{x^2}{8}+\frac{y^2}{4}=1\end{array}\right.,\left\{\begin{array}{l}y=-\frac{1}{2k}x\\ \frac{x^2}{8}+\frac{y^2}{4}=1\end{array}\right.$,由椭圆的对称性可知S四边形ABCD=4×S△AOB=2|OA||OB|sin∠AOB.由此能证明四边形ABCD的面积为定值.

解答 解:(1)∵椭圆$E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1({a>b>0})$的离心率为$\frac{{\sqrt{2}}}{2}$,且过点$({2,\sqrt{2}})$,
∴由题意可得$\left\{\begin{array}{l}\frac{c}{a}=\frac{{\sqrt{2}}}{2}\\ \frac{4}{a^2}+\frac{2}{b^2}=1\\{a^2}={b^2}+{c^2}\end{array}\right.$,解得$\left\{\begin{array}{l}{a^2}=8\\{b^2}={c^2}=4\end{array}\right.$,
∴椭圆的标准方程为$\frac{x^2}{8}+\frac{y^2}{4}=1$.
证明:(2)设A(x1,y1),B(x2,y2),不妨设x1>0,x2>0.
设kAC=k,∵${k_{AC}}•{k_{BD}}=-\frac{b^2}{a^2}=-\frac{1}{2}$,∴${k_{BD}}=-\frac{1}{2k}$
可得直线AC、BD的方程分别为$y=kx,y=-\frac{1}{2k}x$.
联立$\left\{\begin{array}{l}y=kx\\ \frac{x^2}{8}+\frac{y^2}{4}=1\end{array}\right.,\left\{\begin{array}{l}y=-\frac{1}{2k}x\\ \frac{x^2}{8}+\frac{y^2}{4}=1\end{array}\right.$.
解得${x_1}=\frac{{2\sqrt{2}}}{{\sqrt{1+2{k^2}}}},{x_2}=\frac{4|k|}{{\sqrt{1+2{k^2}}}}$.
∴$\overrightarrow{OA}•\overrightarrow{OB}={x_1}{x_2}+{y_1}{y_2}=\frac{1}{2}{x_1}{x_2}=\frac{{4\sqrt{2}|k|}}{{1+2{k^2}}}≤\frac{{4\sqrt{2}|k|}}{{2\sqrt{2}|k|}}=2$,
当且仅当$|k|=\frac{{\sqrt{2}}}{2}$时取等号.
由椭圆的对称性可知S四边形ABCD=4×S△AOB=2|OA||OB|sin∠AOB.
∴$S_{四边形ABCD}^2=4[{{{|{OA}|}^2}{{|{OB}|}^2}-{{({\overrightarrow{OA}•\overrightarrow{OB}})}^2}}]$
=$4[{({x_1^2+y_1^2})({x_2^2+y_2^2})-{{({{x_1}{x_2}+{y_1}{y_2}})}^2}}]$
=$4{({{x_1}{y_2}-{x_2}{y_1}})^2}$=$4{({-\frac{1}{2k}{x_1}{x_2}-k{x_1}{x_2}})^2}$=$4{({k+\frac{1}{2k}})^2}{({\frac{{8\sqrt{2}k}}{{1+2{k^2}}}})^2}$=128,
∴四边形ABCD的面积=$8\sqrt{2}$为定值.

点评 本题考查椭圆方程的求法,考查四边形的面积为定值的证明,是中档题,解题时要认真审题,注意韦达定理、椭圆性质、弦长公式的合理运用.

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