Question

如图已知抛物线y=ax2+bx+3与x轴交于AB两点过A的直线l与抛物线交于点c其中A(1,0)c(4,3)若点E是解析

Answer 1

(1, 0 ) : 0= a+ b + 3

a+b=-3--(1)

(4, 3) : 3= 16a + 4b + 3

4a + b=0 ---(2)

(1), (2) : 4a+(-3 -a ) =0

3a - 3=0

a=1-------> b =-4

y=x^2 - 4x + 3 =(x-1)(x-3)

直线 AC : y=(3-0) /(4-1) (x-1) + 0

= (x-1)

点E, 直线 AC 距离

d= { |t-(t^2 - 4t +3) -1 |} /{ sqrt{1^2 + 1^2 }}

= {| -t^2 +5t -4 |} /{ sqrt{2 } } = {| t^2 -5t +4 |} /{ sqrt{2 } }

= {| (t - 5/2 )^2 +4 - 25/4 |} /{ sqrt{2 } }

因此, E点坐标 (5/2 ,- 3/4)