已知3x加2t=4,2y-t=3
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x=(√3/2)(2t);y=2-(2t)/2,令2t=T,则X=√3T/2,y=2-T/2,则|T|表示直线上任一点到(0,2)的距离,将X=√3T/2,y=2-T/2代入y^2=2x得:(2-T/
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
左边的式子乘以2,右面的式子乘以3,相加.最后y=(5-14x)/3
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
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+
你好:x+y+z=6为①2x-z+t=-2为②y+z+t=4为③x-2y+t=-4为④由(3)加②得2x+2t+y=2(5)把④*2得2x-4y+2t=-8(6)(5)-(6)得到由(2x+2t+y)
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=
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)
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)
削元法:2X=2-5t得到:t=(2-2X)/5,带入3y-2t=X得到:3y-2*(2-2x)/5=x,简化得到:x-5y+6=0
y=2x-8.再问:过程........再答:t=(x-3)/2,y=4((x-3)/2-2,=2x-8
3x+4y=2t.[1]2x+3y=t-1.[2][1]*3:9x+12y=6t[2]*4:8x+12y=4t-4二式相减得:x=2t+4y=(2t-3x)/4=(2t-6t-12)/4=-t-3x
x=3t+1t=(x-1)/3y=2t-1=(2x-5)/3
由y+2=3t得y=3t-2又由2x=3-t得t=3-2x所以y=7-6x