顶点在原点，焦点在x轴上的抛物线截直线y=2x-4所得的弦长ΙABΙ=3√5，求此抛物线方程．
热心网友
顶点在原点、焦点在x轴上的抛物线为：y^2=2px将y=2x-4代入，整理得：2x^2-(p+8)x+16 =0x1+x2 = (p+8)/2, x1*x2 = 8 == (x1-x2)^2=(z1+x2)^2-4*x1x2=(p^2+16p)/4|AB|^2 =(x1-x2)^2+(y1-y2)^2 = (x1-x2)^2 +[2*(x1-x2)]^2== (3√5)^2 = 5*(x1-x2)^2 = (p^2+16p)/4== p = 2, -18因此，抛物线方程为：y^2 = 4x, 或: y^2 = -36x
热心网友
设抛物线的方程是 y^2=2mx (m0)消去x得到 y^2-my-4m=0---y1+y2=m;y1y2=-4m(y1-y2)^2=(y1+y2)^2-4y1y2=m^2+16m|AB|^2=(x1-x2)^2+(y1-y2)=(y1-y2)^2{[(x1-x2)/(y1-y2)]^2+1}=(y1-y2)^2*(1/k^2+1)k=2---(m^2+16m)^2*(1/4+1)=9*5---(m^2+16m)6-9=0---(m^2+16m-3)(m^2+16m+3)=0---m=-8+'-√69 or -8+'-√61所以,抛物线的方程是y^2=2mx.(m=以上的4个不同的值)