Question #141908
The line x=y=z-1 intersects the cone x^2+y^2+z^2+2yz+2zx+2xy=0 at exactly one point
1
Expert's answer
2020-11-02T18:51:18-0500

Given equation of line is

Let

x=y=z1=tx=y=z-1=t


we get

x=ty=tz=1+tx=t\\ y=t\\ z=1+t


Given Equation of cone is

x2+y2+z2+2xy+2yz+2xz=0t2+t2+(1+t)2+2t2+2t(1+t)+2t(1+t)=09t2+6t+1=0x^2+y^2+z^2+2xy+2yz+2xz=0\\\to t^2+t^2+(1+t)^2+2t^2+2t(1+t)+2t(1+t)=0\\\to 9t^2+6t+1=0


(3t+1)2=0t=13\to (3t+1)^2=0\\\to t=-\dfrac{1}{3}


so we get,

x=13y=13z=113=23x=-\dfrac{1}{3}\\ y=-\dfrac{1}{3}\\z=1-\dfrac{1}{3}=\dfrac{2}{3}



So given line intersect with cone at (13,13,23)(\dfrac{-1}{3},\dfrac{-1}{3},\dfrac{2}{3}) .


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