在直角坐标系XOY中点P到两点(-3,0)(3,0)的距离之和等于10 设点P的轨迹为曲线C

C为以(-3,0),(3,0)为焦点,长轴为10的椭圆
a = 10/2 = 5
c = 3,b = 4
椭圆:x²/25 + y²/16 = 1
直线:y = (4/5)(x - 3)
代入椭圆方程,整理得:x² - 3x - 8 = 0
x₁ + x₂ = 3,x₁x₂ = -8
线段的中点M(p,q)
p = (x₁ + x₂)/2 = 3/2
q = (y₁ + y₂)/2 = [(4/5)(x₁ - 3) + (4/5)(x₂ - 3)]/2 = (2/5)(x₁ + x₂ - 6) = (2/5)(3 - 6) = -6/5
M(3/2,-6/5)