Asked by Drake
There are two points A=(2,0) and B=(4,0) and a moving point P=(t,t) on the xy-plane. If the minimum value of line AP+ line BP is b, what is the value of b^2?
Answers
Answered by
Steve
AP = √((t-2)^2 + t^2)
BP = √((t-4)^2 + t^2)
b^2 = (AP+BP)^2
= AP^2 + 2 AP*BP + BP^2
= ((t-2)^2 + t^2)+((t-4)^2 + t^2) + 2√(((t-2)^2 + t^2)*((t-4)^2 + t^2))
= 4(t^2 - 3t + 5 + √(t^4-6t^3+18t^2-24t+16))
min b^2 = 20 when t = 4/3
BP = √((t-4)^2 + t^2)
b^2 = (AP+BP)^2
= AP^2 + 2 AP*BP + BP^2
= ((t-2)^2 + t^2)+((t-4)^2 + t^2) + 2√(((t-2)^2 + t^2)*((t-4)^2 + t^2))
= 4(t^2 - 3t + 5 + √(t^4-6t^3+18t^2-24t+16))
min b^2 = 20 when t = 4/3
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.