题目内容
5.椭圆$\frac{{x}^{2}}{{a}^{2}}$$+\frac{{y}^{2}}{{b}^{2}}$=1(a>b>0),两焦点F1,F2,P为椭圆上一点,若存在$\overrightarrow{P{F}_{1}}$$•\overrightarrow{P{F}_{2}}$=c2,求e的范围.分析 可设P(asinθ,bcosθ),可写出F1,F2的坐标,从而得出$\overrightarrow{P{F}_{1}}•\overrightarrow{P{F}_{2}}={a}^{2}si{n}^{2}θ-{c}^{2}+{b}^{2}co{s}^{2}θ={c}^{2}$,从而可以得到$si{n}^{2}θ=\frac{2{c}^{2}-{b}^{2}}{{a}^{2}-{b}^{2}}$,而根据0≤sin2θ≤1便可得到$\left\{\begin{array}{l}{2{c}^{2}-{b}^{2}≥0}\\{2{c}^{2}-{b}^{2}≤{a}^{2}-{b}^{2}}\end{array}\right.$,进一步便得到$\left\{\begin{array}{l}{3{c}^{2}≥{a}^{2}}\\{2{c}^{2}≤{a}^{2}}\end{array}\right.$,这样即可求出$\frac{c}{a}$的范围,即e的范围.
解答 解:设P(asinθ,bcosθ),F1(-c,0),F2(c,0);
∴$\overrightarrow{P{F}_{1}}=(-c-asinθ,-bcosθ)$,$\overrightarrow{{PF}_{2}}=(c-asinθ,-bcosθ)$;
∴$\overrightarrow{P{F}_{1}}•\overrightarrow{P{F}_{2}}={a}^{2}si{n}^{2}θ-{c}^{2}+{b}^{2}co{s}^{2}θ={c}^{2}$;
∴a2sin2θ+b2(1-sin2θ)=2c2;
∴(a2-b2)sin2θ=2c2-b2;
∴$si{n}^{2}θ=\frac{2{c}^{2}-{b}^{2}}{{a}^{2}-{b}^{2}}$;
∴$0≤\frac{2{c}^{2}-{b}^{2}}{{a}^{2}-{b}^{2}}≤1$;
∴$\left\{\begin{array}{l}{2{c}^{2}-{b}^{2}≥0}\\{2{c}^{2}-{b}^{2}≤{a}^{2}-{b}^{2}}\end{array}\right.$;
∴$\left\{\begin{array}{l}{3{c}^{2}≥{a}^{2}}\\{2{c}^{2}≤{a}^{2}}\end{array}\right.$;
∴$\left\{\begin{array}{l}{\frac{{c}^{2}}{{a}^{2}}≥\frac{1}{3}}\\{\frac{{c}^{2}}{{a}^{2}}≤\frac{1}{2}}\end{array}\right.$;
∴$\frac{\sqrt{3}}{3}≤\frac{c}{a}≤\frac{\sqrt{2}}{2}$;
∴e的范围为$[\frac{\sqrt{3}}{3},\frac{\sqrt{2}}{2}]$.
点评 考查根据sin2θ+cos2θ=1设椭圆上点的坐标,根据点的坐标求向量的坐标,以及向量数量积的坐标运算,a2=b2+c2,椭圆离心率的概念及计算公式.
A. | 2015+2016i | B. | 2015-2016i | C. | -2016+2015i | D. | -2016-2015i |
A. | 63 | B. | 127 | C. | 128 | D. | 255 |