Question

10．參數(shù)方程$\left\{\begin{array}{l}{x=\frac{{3t}^{2}}{1+{t}^{2}}}\\{y=\frac{5-{t}^{2}}{1{+t}^{2}}}\end{array}\right.$（t為參數(shù)）表示的圖形為2x+y-5=0（0≤x＜3）．

Answer 1

分析 由題意求出x的范圍，消去參數(shù)t得答案．

解答 解：由x=$\frac{3{t}^{2}}{1+{t}^{2}}$知，t²=0時x=0，t²≠0時$x=\frac{3}{1+\frac{1}{{t}^{2}}}＜3$，
∴0≤x＜3．
再由$\left\{\begin{array}{l}{x=\frac{{3t}^{2}}{1+{t}^{2}}}\\{y=\frac{5-{t}^{2}}{1{+t}^{2}}}\end{array}\right.$，得$y=\frac{5}{1+{t}^{2}}-\frac{{t}^{2}}{1+{t}^{2}}=\frac{5}{1+{t}^{2}}-\frac{x}{3}$，∴$1+{t}^{2}=\frac{15}{x+3y}$，
則$x=\frac{3（\frac{15}{x+3y}-1）}{\frac{15}{x+3y}}=\frac{15-x-3y}{5}$，整理得：2x+y-5=0（0≤x＜3）．
故參數(shù)方程$\left\{\begin{array}{l}{x=\frac{{3t}^{2}}{1+{t}^{2}}}\\{y=\frac{5-{t}^{2}}{1{+t}^{2}}}\end{array}\right.$（t為參數(shù)）表示的圖形為：2x+y-5=0（0≤x＜3）．

點評 本題考查參數(shù)方程化普通方程，考查計算能力，是基礎(chǔ)題．