Question #141009
The line x=y=z-1 intersects the cone x^2+y^2+z^2+2yz+2zx+2xy=0 at exactly one point.
True or false with correct explanation
1
Expert's answer
2020-10-29T18:32:03-0400

x2+y2+z2+2yz+2zx+2xy=0x^2+y^2+z^2+2yz+2zx+2xy=0


Let x = y = z – 1 = t

t2+t2+(t+1)2+2t(t+1)+2(t+1)t+2t2=0t^2+t^2+(t+1)^2+2t(t+1)+2(t+1)t+2t^2=0

4t2+t2+2t+1+2t2+2t+2t2+2t=04t^2+t^2+2t+1+2t^2+2t+2t^2+2t=0

9t2+6t+1=09t^2+6t+1=0

(3t+1)2=0(3t+1)^2=0

3t+1=0    t=133t+1=0 \implies t=-\frac{1}{3}


Hence x=13,y=13,z=23x=-\frac{1}{3}, y=-\frac{1}{3}, z=\frac{2}{3}

Therefore (13,13,23)(-\frac{1}{3},-\frac{1}{3},\frac{2}{3}) is the only intersection point, q.e.d. Answer: True.


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