题目内容
1.在平面直角坐标系xOy中,已知直线$\left\{{\begin{array}{l}{x=-1+\frac{{\sqrt{5}}}{5}t}\\{y=-1+\frac{{2\sqrt{5}}}{5}t}\end{array}}\right.$(t为参数)与曲线$\left\{{\begin{array}{l}{x=sinθ}\\{y=cos2θ}\end{array}}\right.$(θ为参数)相交于A,B两点,求线段AB的长.分析 直线$\left\{{\begin{array}{l}{x=-1+\frac{{\sqrt{5}}}{5}t}\\{y=-1+\frac{{2\sqrt{5}}}{5}t}\end{array}}\right.$(t为参数),消去参数t化为普通方程.由曲线$\left\{{\begin{array}{l}{x=sinθ}\\{y=cos2θ}\end{array}}\right.$(θ为参数),利用倍角公式可得y=1-2sin2θ,联立解出,再利用两点之间的距离公式即可得出.
解答 解:直线$\left\{{\begin{array}{l}{x=-1+\frac{{\sqrt{5}}}{5}t}\\{y=-1+\frac{{2\sqrt{5}}}{5}t}\end{array}}\right.$(t为参数)化为普通方程:y=2x+1.
由曲线$\left\{{\begin{array}{l}{x=sinθ}\\{y=cos2θ}\end{array}}\right.$(θ为参数),可得y=1-2sin2θ=1-2x2(-1≤x≤1),
联立$\left\{\begin{array}{l}{y=2x+1}\\{y=1-2{x}^{2}}\end{array}\right.$(-1≤x≤1),解得$\left\{\begin{array}{l}{x=-1}\\{y=-1}\end{array}\right.$,或$\left\{\begin{array}{l}{x=0}\\{y=1}\end{array}\right.$,
.∴A(-1,-1),B(0,1),
∴|AB|=$\sqrt{(-1-0)^{2}+(-1-1)^{2}}$=$\sqrt{5}$.
点评 本题考查了参数方程化为普通方程、曲线的交点、两点之间的距离公式,考查了推理能力与计算能力,属于中档题.
| A. | 2-$\sqrt{2}$ | B. | 1 | C. | 2$\sqrt{2}$ | D. | 1+$\sqrt{2}$ |