+4 Points \(A\) and \(B \) are on parabola , \(y=3x^2-5x-3\), and the origin is the midpoint of \(\overline{AB}\). Find the square of the length of \(\overline{AB}\).
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 \begin{align*} AB^2&=(a-b)^2+(3a^2-5a-3-(3b^2-5b-3))^2\ &=a^2+b^2+6(a^2-b^2)+10(a-b)+9\ &=a^2+b^2+10(a-b)+9. \end{align*}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.
