Question

急！已知F1,F2是椭圆C:X²/4+Y²/b²=1（0＜b＜2）的两个焦点,P为椭圆上的点,

若|PF1→+PF2→|＝2√3,∠F1PF2＝60º，求S△PF1F2(是向量PF1+PF2的绝对值)

Answer 1

|PF1→+PF2→| = 2|PO→|, 所以 PO=√3

PF1+PF2 = 2a = 4; pF1^2+PF2^2 -2PF1*PF2*cos∠F1PF2 = F1F2^2
可知 (PF1+PF2)^2-3PF1*PF2 = F1F2^2
4b^2 = 3PF1*PF2 .............1#

继续余弦定理 : PO^2+OF1^2 -2OP*OF1*cos∠POF1 = PF1^2
PO^2+OF2^2 -2OP*OF2*cos∠POF2 = PF2^2
∠POF1+∠POF2 = π 两式相加
2(PO^2+OF1^2) = PF1^2+PF2^2=2*(4-b^2 + 3)
2PF1*PF2 = (PF1+PF2)^2-PF1^2+PF2^2 = 16-14+2b^2 = 2b^2+2....2#

联立1#,2#,可知 3(b^2+1) = 4b^2 所以 b^2 = 3,PF1*PF2=4

S△PF1F2 = PF1*PF2sin,∠F1PF2/2=√3 (刚刚最后一步算错了，以此为准)