Question

已知抛物线的焦点坐标为(-3,0),准线方程为x=3,则抛物线方程为(    )

A.x²+6y²=0                                     B.y²+12x=0

C.y+6x²=0                                      D.y+12x²=0

Answer 1

B

解析:由题意知抛物线的开口向左,且=3,∴所求方程为y²=-12x,即y²+12x=0.故应选B.