Answer to Question #141908 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

Answer 1

Given equation of line is

Let

$x=y=z-1=t$

we get

$x=t\\ y=t\\ z=1+t$

Given Equation of cone is

$x^2+y^2+z^2+2xy+2yz+2xz=0\\\to t^2+t^2+(1+t)^2+2t^2+2t(1+t)+2t(1+t)=0\\\to 9t^2+6t+1=0$

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

so we get,

$x=-\dfrac{1}{3}\\ y=-\dfrac{1}{3}\\z=1-\dfrac{1}{3}=\dfrac{2}{3}$

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