+4 Points A and B are on parabola , y=3x2−5x−3, and the origin is the midpoint of ¯AB. Find the square of the length of ¯AB.
Let A be the point (a,3a2−5a−3) and B be the point (b,3b2−5b−3). Since the origin is the midpoint, a+b=0.
The squared distance between A and B is AB2=(a−b)2+(3a2−5a−3−(3b2−5b−3))2 =a2+b2+6(a2−b2)+10(a−b)+9 =a2+b2+10(a−b)+9.Since a+b=0, replacing b with −a, we get AB2=a2+(10a)+9=a2+10a+9.
We need to find the value of a that minimizes this expression. The expression a2+10a+9 is a quadratic expression in a, whose vertex is at −210=−5. Since a is negative, the expression is minimized when a=−5. Therefore, a=−5, and AB^2 = a^2+10a+9 = (−5)^2+10(−5)+9 = 46.
