题目内容
8.已知曲线C1:$\left\{\begin{array}{l}{x=1+tcosα}\\{y=tsinα}\end{array}$(t为参数),曲线C2:$\frac{x^2}{9}$+$\frac{y^2}{2}$=1.(Ⅰ)写出C1的普通方程与C2的参数方程;
(Ⅱ)过坐标原点O做C1的垂线,垂足为A,P为OA中点,当α变化时,求P点的轨迹的参数方程(以α为参数),并指出它是什么曲线.
分析 (Ⅰ)曲线C1:$\left\{\begin{array}{l}{x=1+tcosα}\\{y=tsinα}\end{array}$(t为参数),消去参数t可得普通方程,利用平方关系可得曲线C2的参数方程.(Ⅱ)由曲线C1的普通方程:xsinα-ycosα-sinα=0,设A(sin2α,-sinαcosα)故当α变化时,P点轨迹的参数方程为:$\left\{\begin{array}{l}{x=\frac{1}{2}si{n}^{2}α}\\{y=-\frac{1}{2}sinαcosα}\end{array}\right.$(α为参数),利用倍角公式与平方关系化简可得P点轨迹的普通方程.
解答 解:(Ⅰ)曲线C1:$\left\{\begin{array}{l}{x=1+tcosα}\\{y=tsinα}\end{array}$(t为参数),消去参数t可得普通方程:xsinα-ycosα-sinα=0,
由曲线C2:$\frac{x^2}{9}$+$\frac{y^2}{2}$=1可得参数方程:$\left\{\begin{array}{l}{x=3cosθ}\\{y=\sqrt{2}sinθ}\end{array}\right.$(θ为参数).
(Ⅱ)由曲线C1的普通方程:xsinα-ycosα-sinα=0,设A(sin2α,-sinαcosα),
故当α变化时,P点轨迹的参数方程为:$\left\{\begin{array}{l}{x=\frac{1}{2}si{n}^{2}α}\\{y=-\frac{1}{2}sinαcosα}\end{array}\right.$(α为参数),P点轨迹的普通方程为$(x-\frac{1}{4})^{2}$+y2=$\frac{1}{16}$.
故P点轨迹是圆心为$(\frac{1}{4},0)$,半径为$\frac{1}{4}$的圆.
点评 本题考查了参数方程与普通方程的互化、直线参数方程的应用、中点坐标公式、倍角公式与平方关系,考查了推理能力与计算能力,属于中档题.
| A. | M∈a,a∈α | B. | M∈a,a?α | C. | M?a,a?α | D. | M?a,a∈α |
| A. | $\frac{e}{2}$ | B. | $\frac{\sqrt{e}}{2e}$ | C. | $\frac{2e}{3}$ | D. | e |
| A. | f(x)=x2+x+1 | B. | f(x)=x2-x-2 | C. | f(x)=x2-x+1 | D. | f(x)=x2+x-2 |
| A. | $\sqrt{2}$ | B. | 2 | C. | $\sqrt{2}$+$\sqrt{2}$i | D. | $\sqrt{2}$-$\sqrt{2}$i |