第1个回答 2013-08-11
解:根据题意得
配方得:(x-t-3)^2+(y+1-4t^2)^2 =-(7t+1)(t-1)
-(7t+1)(t-1)>0
-1/7<t<1
配方:
[x-(t+3)]^2+[y+(1-4t^2)]^2=(t+3)^2+(1-4t^2)^2-16t^4-9
[x-(t+3)]^2+[y+(1-4t^2)]^2=-7t^2+6t+1
[x-(t+3)]^2+[y+(1-4t^2)]^2=-7(t^2-3/7)^2+16/7
因此r^2<=16/7
即半径取值范围是(0,4/√7]本回答被提问者采纳