不妨设a>b
M(acost,bsint)
B1(0,b).B2(0,-b)
MB1:y-b=(bsint-b)x/(acost),--> P(acost/(1-sint),0)
MB2:y+b=(bsint+b)x/acost)--> Q(acost/(1+sint),0)
|OP|•|OQ|=|(acost)/(1-sint)*(acost)/(1+sint)|
=a^2(cost)^2/[1-(sint)^2]
=a^2
所以得证.
不妨设a>b
M(acost,bsint)
B1(0,b).B2(0,-b)
MB1:y-b=(bsint-b)x/(acost),--> P(acost/(1-sint),0)
MB2:y+b=(bsint+b)x/acost)--> Q(acost/(1+sint),0)
|OP|•|OQ|=|(acost)/(1-sint)*(acost)/(1+sint)|
=a^2(cost)^2/[1-(sint)^2]
=a^2
所以得证.