已知方程x² y²-2(t-3)x-2(1-4t²)y 16t的四次方 9=0

x=y^2-y-2再问：求解答过程再答：y=t-1,t=y+1,代入，x=(y+1)^2-3(y+1)+1=y^2+2y+1-3y-3+1=y^2-y-1检验的时候发现上面回答的错了，答案是y^2-y

将P（3,4t^2）带入x^2+y^2-2(t+3)x+2(1-4t^2)y+16t^4+9

x-1=(t+1)/(t-1)-1=2/(t-1)t-1=2/(x-1)t=(x+1)/(x-1)t^2+t+1=(x+1)^2/(x-1)^2+(x+1)/(x-1)+1=(3x^2+1)/(x-1

由题得：直线L的斜率k=tan60°=根号3所以,直线L的普通方程：y-3=(根号3)(x-1)因为,y-3=[(根号3)/2]t-------------------(1)x-1=(1/2)t---

x=t^2+1t^2=x-1t=根号(x-1)y=4t-t^2=4根号(x-1)-x+1y=4根号(x-1)-x+1(x>=1)

1)x^2+y^2-2(t+3)x+2(1-4t^2)y+16t^4+9=0[x-(t+3)]^2+[y+(1-4t^2)]^2=-7t^2+6t+1R^2=-7t^2+6t+1-7t^2+6t+1>

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

①x2+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+

圆心坐标为（-t-3,1-4t^2）根据参数方程x=-t-3;y=1-4t^2;解得y=-4x^2-24x-35,x∈（-4,-20/7）;再问：圆心坐标应为（t+3,4t^2-1)

方程化为：[x-（t+3）]²+[y+（1-4t²）]²＝-8（t-3/8）²+9/8当t＝3/8时,圆面积＝9π/8最大.

x+y-2（t+3）x+2（1-4t）y+16t^4+9=0(x-(t+3))+(y+(1-4t))+16t^4+9=(t+3)+(1-4t)(x-(t+3))+(y+(1-4t))+16t^4+9=

-4t=2x-6=y+12x-y-7=0

这么晚不睡写作业好孩子啊,说下思路吧,既然表示的是圆,那么x.y前的系数就要有范围,不过我记不太清楚了,你看看公式吧再问：t求出了=3/7，可是求不出圆点的坐标，怎么具体解再答：再问：原来有公式，我还

设为一个新的参数t,两个t不一样.2/根号5是直线cos倾斜角1/根号5是sin将x=1+2/根号5t和y=2+1/根号5t里的xy代入x^2+y^2=9得到一个含t的二元一次方程,用韦达定理,求（t

1.方程可化为：[x-(t+3)]^2+[y+(1-4t^2)]^2=(t+3)^2+(1-4t^2)^2-(16t^4+9)因为该方程表示圆,故(t+3)^2+(1-4t^2)^2-(16t^4+9

因为x2+y2-2(t+3)x+2(1-4t2)y+16t4+9=0所以(x-t-3)^2+(y+1-4t^2)=(t+3)^2+(1-4t^2)^2-16^4-9=1+6t-7t^2所以当半径最大时

－1/7＜t＜1

圆的半径为r=1/2*√D^2+E^2-4F代入讨论即可很高兴为您解答,【学习宝典】团队为您答题.请点击下面的【选为满意回答】按钮,

四个①x-2=3/x;②0.4x=11；③x/2=5x-1；⑤t=0；再问：x/2=5x-1也是啊？再答：是啊，x的次数只有1你是写x分之2还是2分之xx分之2就不是了2分之x就是了

配方：[x-(t+3)]²+[y+(1-4t²)]²=(t+3)²+(1-4t²)²-16t^4-9[x-(t+3)]²+[y+(