Question

若P∈(x^2)/(a^2)+(y^2)/(b^2)=1,F1,F2为焦点,且PF1⊥PF2,则|PF1-PF2|=__

Answer 1

由椭圆的定义

|PF1|+|PF2|=2a

|F1F2|=2c

PF1⊥PF2

则|PF1|^2+|PF2|^2=|F1F2|^2=4c^2

记|PF1|=x |PF2|=y

x+y=2a

x^2+y^2=4c^2

(x+y)^2=4a^2

x^2+y^2+2xy=4a^2

xy=2a^2-2c^2

|x-y|^2=x^2+y^2-2xy=4c^2-4a^2+4c^2=8c^2-4a^2

|x-y|=2√(2c^2-a^2)=2√(a^2-2b^2)