Prove all tangent plane to the cone $x^2+y^2=z^2$ goes through the origin

Question

Good morning, i need help with this exercise.

Prove all tangent plane to the cone $x^2+y^2=z^2$ goes through the origin

My work:

Let $f:\mathbb{R}^3\rightarrow\mathbb{R}$ defined by $f(x,y,z)=x^2+y^2-z^2$

Then,

$\nabla f(x,y,z)=(2x,2y,-2z)$

Let $(a,b,c)\in\mathbb{R}^3$ then $\nabla f(a,b,c)=(2a,2b,-2c)$

By definition, the equation of the tangent plane is

\begin{eqnarray} \langle(2a,2b,-2c),(x-a,y-b,z-c)\rangle &=& 2a(x-a)+2b(y-b)+2c(z-c)\\ &=&2ax-2a^2+2by-2b^2+2cz-2c^2 \\ &=&0 \end{eqnarray}

In this step i'm stuck, can someone help me?

Answer 1

The equation for the plane should be $2a(x-a)+2b(y-b)-2c(z-c)=2ax-2a^{2}+2by-2b^{2}-2cz+2c^{2}=0$. Now the point $(a,b,c)$ lies on the cone, so $a^{2}+b^{2}-c^{2}=0$, so simplifying the equation for the plane, we have then $2ax+2by-2cz=0$ and this equation goes through the origin since $2a\cdot 0+2b\cdot 0-2c\cdot 0=0$.

Answer 2

The equation of the tangent plane is

$$z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$

For the cone we have

$$f(x,y)=\sqrt{x^2+y^2}\implies f_x=\frac{x}{\sqrt{x^2+y^2}} \quad f_y=\frac{y}{\sqrt{x^2+y^2}}$$

Thus the tangent plane at $(x_0,y_0,z_0)$ is $$z-z_0=\frac{x_0}{z_0} (x-x_0)+\frac{y_0}{z_0}(y-y_0)$$ $$z\cdot z_0-z_0^2=x\cdot x_0-x_0^2+y\cdot y_0-y_0^2$$ $$z\cdot z_0=x\cdot x_0+y\cdot y_0$$

Answer 3

Actually you want $(a, b, c)$ on the cone... so $a^2+b^2-c^2=0$...

Now the equation of the tangent plane is always satisfied by the origin $(0, 0, 0) $

Note: you lost a minus sign in your equation (the $z $ -term). ..

Substitute $(x, y, z)=(0, 0, 0) $ to find that the origin satisfies the equation of the plane...