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已知方程x2+y2-2(t+3)x+2(1-4t2)y+16t2+9=0(t属于R)的图形是圆
1.求t的取值范围
2.求其中面积最大的圆的方程
人气:359 ℃ 时间:2020-04-04 06:05:15
解答
(1)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
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