# Show that the tangent plane of the cone $z^2=x^2+y^2$ at (a,b,c)$\ne$0 intersects the cone in a line

My attempt: Let $$f(x,y,z)=x^2+y^2-z^2$$ and calculate the total derivative $$Df(a,b,c)=(2a,2b,-2c)$$. The tangent plane is $$g(x,y,z)=f(a,b,c)+Df(a,b,c)(x-a,y-b,z-c)=2ax+2by-2cz-a^2-b^2+c^2$$ Then let $$g(x,y,z)=f(x,y,z)$$ to find the intersction got $$(x-a)^2+(y-b)^2-(z-c)^2=0$$ Does this equation means the tangent plane intersects the cone in a line?

• That last equation is just that of the cone translated so that its vertex is at $(a,b,c)$. – amd May 20 at 7:34

You took a point $$(a,b,c)$$ belonging to the cone $$C \equiv z^2=x^2+y^2$$. So you must have $$a^2+b^2=c^2$$ and the equation of the tangent plane simplifies to $$2ax+2by-2cz=0$$.
From there, it is easy to see that for all points $$P_\lambda = (\lambda a, \lambda b, \lambda c)$$ that belongs to a line included in $$C$$, the equation of the tangent plane at $$P_\lambda$$ is also $$2ax+2by-2cz=0$$. That proves the expected resut.