Question

设抛物线y^2=2x与y=x+b想交于AB两点.弦AB长12求b的值

Answer 1

y^2=2x y=x+b 联立：x^2+2(b-1)x+b^2=0 x1+x2=2(1-b) x1x2=b^2 (x1-x2)^2=(x1+x2)^2-4x1x2=4(b-1)^2-4b^2=4-8b |x1-x2|=根号(4-8b)=12/根号2 1-2b=18 b=-17/2.