Question

高中数学微积分题目

有一条直线与抛物线y=x^2相交于A、B两点，线段AB与抛物线所围成的图形面积恒等于4/3，求线段AB的中点P的轨迹方程。

Answer 1

Let a be the x-coordinate of point A, b be the x-coordinate of point B.
then the points A, B have coordinates (a, a²)(b, b²)

Area between segment AB and the parabola is
[area of the trapezoid] - ʃ [a~b] x² dx = 4/3
which is: (a²+b²)(b-a)/2 - b³/3 + a³/3 = 4/3
which can be simplified into: a²b/2 - a³/6 + b³/6 - ab²/2 = 4/3
which can be factorized into: -(a-b)³/6 = 4/3
solve the equation, we get b-a=2

The coordinates of point P is (a,a²)(b,b²)/2 = ((a+b)/2, (a²+b²)/2)
Substitute b-a=2 into ((a+b)/2, (a²+b²)/2), we get (b-1, b²-2b+2)
which is P(b-1)=b²-2b+2
Let m be the x-coordinate of point P, then m=b-1, b=m+1
P(m)=(m+1)²-2(m+1)+2
P(m)=m²+1
or y=x²+1

做人要厚道。。这种题还是提高一下悬赏分的比较好。。谢谢