Question

20．曲線$\left\{\begin{array}{l}{x=t+\frac{1}{t}}\\{y=t-\frac{1}{t}}\end{array}\right.$（t為參數(shù)）的普通方程為$\frac{{x}^{2}}{4}$-$\frac{{y}^{2}}{4}$=1．

Answer 1

分析 將參數(shù)方程的兩式平方相減即可消去參數(shù)，得到普通方程．

解答 解：∵$\left\{\begin{array}{l}{x=t+\frac{1}{t}}\\{y=t-\frac{1}{t}}\end{array}\right.$，∴$\left\{\begin{array}{l}{{x}^{2}={t}^{2}+\frac{1}{{t}^{2}}+2}\\{{y}^{2}={t}^{2}+\frac{1}{{t}^{2}}-2}\end{array}\right.$，∴x²-y²=4，即$\frac{{x}^{2}}{4}-\frac{{y}^{2}}{4}=1$．
故答案為：$\frac{{x}^{2}}{4}-\frac{{y}^{2}}{4}=1$．

點評 本題考查了參數(shù)方程與普通方程的轉(zhuǎn)化，屬于基礎題．