Answer to Question #124973 in Analytic Geometry for Nikhil

Question #124973
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
2020-07-05T16:35:46-0400

Let "x=y=z-1=t" hence

"x^2+y^2+z^2+2yz+2zx+2xy=0\\Leftrightarrow\\\\\nt^2+t^2+(t+1)^2+2t(t+1)+2(t+1)t+2t^2=0\\Leftrightarrow\\\\\n4t^2+(t^2+2t+1)+4(t^2+t)=0\\Leftrightarrow\\\\\n9t^2+6t+1=0\\Leftrightarrow\\\\\n(3t+1)^2=0\\Leftrightarrow\\\\\n3t+1=0\\Leftrightarrow\\\\\nt=-\\frac13"

hence "x=t=-\\frac13,\\,y=t=-\\frac13,\\,z=t+1=\\frac23" therefore "\\left(-\\frac13,-\\frac13,\\frac23\\right)" is the only intersection point, QED.


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