Question

设A,B为抛物线y=-3x＾2-2x+k图像与X轴两个交点,M为抛物线顶点,当ΔMAB为等腰直角三角形时,求k值.

Answer 1

因抛物线与x轴有两个相异的交点,

△=4-4k×（-3）＞0,

得k＞- 1／3,

∵∠AMB=90°,AM=BM,过M作MN⊥x轴于N,

∴MN= 1／2AB,

∵MN= ﹙4k×(-3)-4﹚／4×(-3)=k+ 1／3,

AB=√ (x1-x2)²,

= √(x1-x2)²-4x1x2,

= √(-2／3)2-4(-k／3),

= 2／3√﹙1+3k﹚,

∴k+ 1／3= 1／2× 2／3√﹙1+3k﹚．

解k1=0,k2=- 1／3（舍去）．

k=0