Question

抛物线y=-1/3(x-m)*2+k的顶点在抛物线y=x*2上,且在X轴上截得线段长是4根号3,求抛物线的解析式

Answer 1

由y=-1/3(x-m)^2+k得抛物线顶点M(m,k)

展开得y=-(1/3)x^2+(2/3)mx-(1/3)m^2+k

所以抛物线与x轴的两个交点是方程-(1/3)x^2+(2/3)mx-(1/3)m^2+k=0的两个解

设两个解分别是x1,x2

则x1+x2=-b/a=-(2/3)m/(-1/3)=2m

x1*x2=c/a=(-(1/3)m^2+k)/(-1/3)=m^2-3k

(x1-x2)^2=(x1+x2)^-4*x1*x2=12k=(4根号3)^2=48 得出 k=4

因为顶点M(m,k)在抛物线y=x^2上,所以k=m^2,因为k=4,所以m=2或-2

k=4,m=2或-2