Analysis: ∵ parabola y 2 = x, its focus f (1/4,0), and its upper point P(x, y).

Distance from p to a (3,-1) | pa | = √ [(x-3) 2+(y+1) 2)]

Distance from p to f (1/4,0) | pf | = √ [(x-1/4) 2+y 2)]

This question is endless,

Analysis: ∫ x 2/3+y 2/4 =1,a 2 = 4, b 2 = 3, c 2 =1.

When p coincides with the left and right vertices of the ellipse, ∠F 1PF2 is the largest.

At this time, pf 1 = pf2 = a = 2, and f 1f2 = 2c = 2.

⊿F 1PF2 is an equilateral triangle.

∴∠F 1PF2=π/3

Option c