如何用初三数学知识证明抛物线定义成立?答:证明:不妨设抛物线是x^2=4py(p>0),准线是y=-p,焦点F(0,p)设M(t,-p)是准线上任意一点,过M作抛物线的两条切线MA、MB,A、B是切点。因A、B在抛物线上,设A(2pm,pm^2),B(2pn,pn^2) (m≠n)由x^2=4py 得y=x^2/(4p), y'=x/(2p)在A处切线斜率k=m,切线方程是mx-y-pm^...
初三数学抛物线答:点(0, -3)和(1, 0)在抛物线上:a + k = -3 4a + k = 0 解得a = 1, k = -4 抛物线为y = (x + 1)^2 - 4 = x^2 + 2x - 3 = (x - 1)(x + 3)x = -3, y = m = 0 ak + h = 1*(-1)-4 = -5 (2) A(-3, 0), B(1, 0), C(0, -3), D...