First, substitute x= 1y=2, and get the equation b c= 1. So y=x2 bx 1-b The ordinate of the parabola vertex is -b2/4-b 1 (minus b squared divided by 4 minus b plus 1). If the two parts of the equation x2 bx 1-b=0 are set as x 1, x2, then Vader's theorem | BC |=| X2-X 1 | = √ δ = √ B24B-4, then (√ b24b-4) √ 3/2 = |-. Because the parabola has an upward opening and two intersections with the X axis, -b2/4-b 1 is less than zero, the equation becomes (√b2 4b-4)√3/2=b2/4 b- 1, that is, (√ b24b-4) √ 3/2 = (b2 Divide √b2 4b-4 by 2√3= √b2 4b-4 on the right, and B24B-4 on both sides =12. Solve this equation to get b =-2 2 √ 5. Verify that δ > 0 is consistent with the meaning of the question. So b =-2 2 √ 5.