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(x^2)+(y^2)-2(t+3)x+2(1-4t^2)y+(16t^4)+9=0
1.t为何值时,方程表示圆?
2.t为何值时,半径R的值大?
人气:228 ℃ 时间:2020-03-19 01:10:45
解答
x2+y2-2(t+3)x+2(1-4t2)y+16t4+9=0
[x-(t+3)]^2+[y+(1-4t^2)]^2=-16t^4-9+(t+3)^2+(1-4t^2)^2
则-16t^4-9+(t+3)^2+(1-4t^2)^2〉0
-16t^4-9+t^2+6t+9+1-8t^2+16t^4>0
-7t^2+6t+1>0
7t^2-6t-1<0
(t-1)(7t+1)<0
-1/7当t=3/7时,R最大
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