Question

15．已知抛物线y²=4x，直线y=x-1，求直线与抛物线的交点坐标．

Answer 1

分析 联立方程得$\left\{\begin{array}{l}{{y}^{2}=4x}&{①}\\{y=x-1}&{②}\end{array}\right.$，利用代入消元法即可解出．

解答 解：联立方程得$\left\{\begin{array}{l}{{y}^{2}=4x}&{①}\\{y=x-1}&{②}\end{array}\right.$，
将②代入①（1）得：x²-6x+1=0，
解得：$x=3±2\sqrt{2}$，
代入②得：$y=2±2\sqrt{2}$，
∴$\left\{\begin{array}{l}{x=3+2\sqrt{2}}\\{y=2+2\sqrt{2}}\end{array}\right.$或$\left\{\begin{array}{l}{x=3-2\sqrt{2}}\\{y=2-2\sqrt{2}}\end{array}\right.$
∴直线与抛物线的交点坐标为$（3+2\sqrt{2}，2+2\sqrt{2}）$或$（3-2\sqrt{2}，2-2\sqrt{2}）$．

点评 本题考查了代入消元法、直线与抛物线的交点坐标求法，考查了推理能力与计算能力，属于中档题．