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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?

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

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  • $\begingroup$ Thank!! I got it. $\endgroup$ – Steven Lu May 20 at 7:28

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