how can you tell by seeing $2ax+2by-2cz=0$ that every tangent plane passes through the origin?

Question

I am trying to solve this problem that asks to show every plane that is tangent to the cone $x^2+y^2=z^2$ passes through the origin.

I saw other questions related because there were already some questions like this. But what I am wondering is something different.

It goes as "set $f(x,y,z)=x^2+y^2-z^2$ then $$\nabla f(x,y,z)= <2x,2y,-2z>$$ Equation of tangent at point $P=(a,b,c)$ is $$\nabla f(a,b,c)=<2a,2b,-2c>$$ $$<2a,2b,-2c> \cdot <x-a,y-b, z-c> = 0$$ $$ 2ax-2a^2+2by-2b^2+(-2cz+2c^2)=0$$ $$ 2ax-2a^2+2by-2b^2-2cz+2c^2=0$$ $$2ax+2by-2cz-(2a^2+2b^2-2c^2)=0$$ $$2ax+2by-2cz=0$$

Now what I wonder is how the last equation $2ax+2by-2cz=0$ says "every tangent plane to the cone passes through the origin"? How can people by seeing this equation gather that every tangent plane passes through the origin?

Answer 1

Hint: Plug $x=y=z=0$ in your equation and you will see it.

Answer 2

Equations "carve out" sets of points, in the right context. Viewed as a restriction on points $(x,y)$, you should be familiar with the fact that $x^2+y^2=1$ carves out a circle. How? Well, all the points $(x,y)$ which satisfy the equation, when you plot them, make a circle. We can view this in another way: the points whose distance from the origin is 1. Now that carves out a circle, for sure! But distance from the origin is:

$$\text{distance between $(0,0)$ and $(x,y)$}=\sqrt{(x-0)^2+(y-0)^2}$$

So the sentence "the distance from the origin is 1" is

$$1=\sqrt{x^2+y^2}\iff 1=x^2+y^2$$

Equations carve out sets of points. If you satisfy the equation, you're in. If you don't, your're not.

Notice that the point $(0,0,0)$ satisfies the equation $2ax+2by-2cz=0$. So, since we know that $2ax+2by-2cz=0$ carves out a plane and we know that $(0,0,0)$ satisfies the equation, we know that $2ax+2by-2cz=0$ goes through the origin.

In the same way, the circle above "goes through" the point $(1,0)$.

Answer 3

This being a cone, for every point $p$ of the cone (other than the vertex) the ray from the vertex to that point is in the cone. The tangent plane at $p$ must include that ray, and thus passes through the vertex of the cone.