Math
posted by Drake on .
There are two points A=(2,0) and B=(4,0) and a moving point P=(t,t) on the xyplane. If the minimum value of line AP+ line BP is b, what is the value of b^2?

AP = √((t2)^2 + t^2)
BP = √((t4)^2 + t^2)
b^2 = (AP+BP)^2
= AP^2 + 2 AP*BP + BP^2
= ((t2)^2 + t^2)+((t4)^2 + t^2) + 2√(((t2)^2 + t^2)*((t4)^2 + t^2))
= 4(t^2  3t + 5 + √(t^46t^3+18t^224t+16))
min b^2 = 20 when t = 4/3