Given that P is any point on the ellipse x ^ 2 / 4 + y ^ 2 = 1, F1 and F2 are the two focuses of the ellipse, find the minimum value of absolute value Pf1 ^ 2 + absolute value PF2 ^ 2

PF1^2+PF2^2=(PF2+PF2)^2-2PF1PF2>=
=(PF2 + PF2) ^ 2 - (PF2 ^ + PF2 ^ 2), so 2 (Pf1 ^ 2 + PF2 ^ 2) > 2=
=(PF2+PF2)^2 =4a^2,
(Pf1 ^ 2 + PF2 ^ 2) > = 2A ^ 2 = 8, if and only if PF2 = PF2 = a, take the equal sign

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