Question

已知抛物线

已知抛物线y=六分之一（x-2）（x-2t-3）(t>0)与x轴交与点A,B（点A在点B得左边），与y轴交与点C
①求A,B,C各点的坐标，可用含t的代数式表示
②设△ABC的面积为二分之二十一，求抛物线的解析式

Answer 1

(1)
y= (1/6)(x-2)(x-2t-3)
y =0
=> x= 2 or 2t+3
A(2t+3,0), B(2,0)
x=0
y=(1/6)(-2)(-2t-3)=(1/3)(2t+3)
C(0,(1/3)(2t+3))
(2)
|AB| = 2t+1
|OC| = (1/3)(2t+3)
△ABC的面积= 21/2 = (1/2)|AB||OC|
=>21/2 = (1/6)(2t+3)(2t+1)
63= 4t^2+8t+3
t^2+2t-15 =0
(t+5)(t-3)=0
t= 3 or -5(rejected)
y= (1/6)(x-2)(x-2t-3)
= (1/6)(x-2)(x-9)

Answer 2

(x-2)(x-2t-3)/6=0 xa=2 xb=2t+3 当x=0 时 y=(2t+3)/3
(xb-xa)(2t+3)/3=21 解得t=2t+1=7 t=3 2t+1=-9 t=-4
所以 y=(x-2)(x-9)/6 或 y=(x-2)(+5)/6