Answer
We verify this by using Theorem 5 to evaluate the tangent plane to the cone.
Work Step by Step
Let $F\left( {x,y,z} \right) \equiv {x^2} + {y^2} - {z^2}$.
So, we can write the equation of the cone as
$F\left( {x,y,z} \right) \equiv {x^2} + {y^2} - {z^2} = 0$
The gradient of $F\left( {x,y,z} \right)$ is
$\nabla F = \left( {{F_x},{F_y},{F_z}} \right) = \left( {2x,2y, - 2z} \right)$
Let $P = \left( {a,b,c} \right)$ be any point on the cone other than the origin, that is, $\left( {a,b,c} \right) \ne \left( {0,0,0} \right)$. Then by Theorem 5, the tangent plane to the cone at $P$ has equation
${F_x}\left( {a,b,c} \right)\left( {x - a} \right) + {F_y}\left( {a,b,c} \right)\left( {y - b} \right) + {F_z}\left( {a,b,c} \right)\left( {z - c} \right) = 0$
$2a\left( {x - a} \right) + 2b\left( {y - b} \right) - 2c\left( {z - c} \right) = 0$
Divide both sides by $2$ gives
$a\left( {x - a} \right) + b\left( {y - b} \right) - c\left( {z - c} \right) = 0$
It can be simplified to
$ax + by - cz = {a^2} + {b^2} - {c^2}$
Since $P = \left( {a,b,c} \right)$ is a point on the cone, we have ${a^2} + {b^2} - {c^2} = 0$. So, the equation of the tangent plane becomes
$ax + by - cz = 0$
By assumption $\left( {a,b,c} \right) \ne \left( {0,0,0} \right)$, so the trivial solution is the origin $\left( {x,y,z} \right) = \left( {0,0,0} \right)$. Since $P = \left( {a,b,c} \right)$ is arbitrary, we conclude that every tangent plane to the cone ${x^2} + {y^2} - {z^2} = 0$ passes through the origin.
