1. The problem statement, all variables and given/known data If the normal to the parabola y^2 = 4ax at the point (at^2 , 2at) cuts the parabola again at (aT^2, 2aT), then minimum value of T^2 is ans: 8 I got the answer but I don't know why it should be the answer? 2. Relevant equations Equation of normal to the parabola in parametric form can be written as y-2at = -t(x-at^2) 3. The attempt at a solution So, in the above equation, I substituted (aT^2, 2aT) as this point will lie on the normal. 2aT - 2at = -t(aT^2 - at^2) On simplifying, t^2 + tT + 2 = 0 From here on, I don't quite understand why what I did works. Minimum value of this quadratic will be at (-b/2a, f(-b/2a) ). i.e., (-T/2, f(-T/2) ) f(-T/2)= (- T^2 / 4 ) + 2 I equate ( -T^2 / 4 ) + 2 to 0. I get T^2 = 8. What did I just do?