证明：设椭圆方程：(x²/a²)+(y²/b²)=1.(a>b>0).点P(acost,bsint),(t∈R),由对称性,不妨设焦点为F(c,0).则|PF|²=(acost-c)²+(bsint)²= (a-ccost)².===>|PF|=a-ccost.∴|PF|max=a+c,|PF|min=a-c.