x=3-2t ,y=-1-4t
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x=1-2t,2t=1-xy=(5-2t)/(7-4t)=[5-(1-x)]/[7-2(1-x)]=(4-x)/(5-2x)x=(5y-4)/(2y-1)
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
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>
由x=1-t/2+t得t=2(x-1)将其代入y=5+4t/3-2t得y=5-4(x-1)/3
x=2t/(1+t)x+xt=2tt=x/(2-x)代入y,得:y=x/(2-x)/[1-x/(2-x)]=x/[2-x-x]=x/(2-2x)再问:看不懂,请详细介绍再答:从第一个方程先解出t,它是
答案:y=2x+3x=2t,y=4t+3因为2x=4t4t+3=y
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+
t=+1,t=-1斜率都是无限,怀疑你的图像画错了
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
x=(1-t)/(1+t)1-t=x(1+t);t=(1-x)/(x+1);将t代入y代数式,化简后得出y=(5x+1)/(5x-1)
1.x^2+y^2-2(t+3)x+2(1-4t^2)y+16t^4+9=0,配方得[x-(t+3)]^2+[y+1-4t^2]^2=(t+3)^2+(1-4t^2)^2-16t^4-9=1+6t-7
这是参数方程求导dy/dx=(dy/dt)(dt/dx)=(dy/dt)/(dx/dt)=(t^3-3t)`/(3t^4+6t)`=(3t^2-3)/(12t^3+6)
有所谓的拉格朗日乘数法可以解决这个多元函数求极值问题设L(x,y,t,u)=5(x-y)^2+4y^2+3t^2+u(2x+y+t-8)分别对x,y,t,u求偏导再令其等于0,得4个方程:L'x=10
[x-(t+3)]²+[y+(1-4t²)]²=-16t^4-9+(t+3)²+(1-4t²)²是圆则r²=-16t^4-9+(t