Question

椭圆x^2/a^2+y^2/b^2=1(a>b>0)上两点P(x,y)Q(x',y'),证明不等式a^2+b^2>=(x

Answer 1

设x=acosp,y=bsinp

x+y=acosp+bsinp=根号(a^2+b^2)sin(p+q)

(x+y)^2=(a^2+b^2)*[sin(p+q)]^2

因为[sin(p+q)]^2