Answer to Question #141009 in Analytic Geometry for Dhruv Rawat

Question

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

Answer 1

$x^2+y^2+z^2+2yz+2zx+2xy=0$

Let x = y = z – 1 = t

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

$4t^2+t^2+2t+1+2t^2+2t+2t^2+2t=0$

$9t^2+6t+1=0$

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

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

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

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