X=t²且Y=t³,解成普通方程-搜答案-拍题作业网

x=t+1/t,y=t-1/tx^2-y^2=4

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

3X+5Y-11=0由X=－3+（5/T+1）Y=4－（3/T+1）两式得

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>

用X=1+4/5t求t：t=5/4(x-1)再带入到y里面：y=1-3/4(x-1)=-3/4x+7/4好啦

x+y=2/tx-y=2tx+y=4/(x-y)x^2-y^2=4

X=t+1/tY=t-1/t都平方得X^2=t^2+2+1/t^2,Y^2=t^2-2+1/t^2X^2-Y^2=4

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

（1）由均值不等式可得x=t^2+1/t^2>=2,所以x^2-y^2=(x+y)(x-y)=(2t^2)*(2/t^2)=4（x>=2）.它表示中心在原点,焦点在x轴上,实轴长为4的双曲线的右支.（

x+y=2tx-y=2/t(x+y)*(x-y)=4

x=t+1/ty=t-1/t两式相加得：t=(x+y)/2代入其中1式得：x=(x+y)/2+2/(x+y)化为：x^2-y^2=4此为双曲线.

第一个应该是或,也就是并集max是较大的那个,也就是X,Y至少一个大於t第二个是两个同时满足,属於交集min是较小的哪个,较小的都大於t那麼X,Y都大於t了当P(X>t)P(Y>t)=P(X>t,Y>

x/y=1/t,t=y/x代入得：x=3y/x/(1+y^2/x^2)x^2=3yx^2/(x^2+y^2)x^2+y^2=3y此为一个圆.

先问一下楼主：x=-t²+1/t²+1还是x=-t²+1/(t²+1)?y=t/t²+1还是y=t/(t²+1)?OKx²=t&s

（1）当t=1时,方程对应一个点（2,2）（2）当t≠1时,由y=t+1/t两边×t整理得t²=ty-1,代入x=t²-2t+3中得：x=ty-2t+2解得t=(x-2)/(y-2

y=(1-x)*4/3

x²=(1-2t²+t²*t²)/(1+t²)²=1-4t²/(1+t²)²(1)4y²=4t&su

将y=t^2-t-1方程两边同乘以2,得2y=2t^2-2t-2,与x=2(t^2)-t-3方程相减,得t=x-2y+1,将t=x-2y+1代入x=2(t^2)-t-3,y=t^2-t-1其中一个方程

y/x=t把t=y/x代入x=2/(t²+1)得：x=2/(y²/x²+1)x=2x²/(x²+y²)x³+xy²-2x

X=T+1/T(T+1/T)X=(T+1/T)^2T^2+1/T^2=(T+1/T)X-2Y=(T+1/T)X-2