已知抛物线解析式:y=ax2+bx+c.如何求焦点坐标,和准线的解析式.已知抛物线焦点和准线,怎么求解析式.初三数学,别弄得太高深..
y = ax², x² = y/a = 2py
p = 1/(2a)
y = ax²的焦点为(0, 1/(4a)), 准线y = -1/(4a)
y = ax² + bx + c是y = ax²向左平移b/(2a), 向上平移c - b²/(4a)得到的;焦点和准线平移后变为:
F(-b/(2a), 1/(4a) +c - b²/(4a))
y = -1/(4a) + c - b²/(4a) = c - (1+ b²)/(4a)