# Tangent plane to double sided cone

• Jan 28th 2009, 05:58 PM
Pur
Tangent plane to double sided cone
Given $\displaystyle z^2=x^2+y^2$, show that every tangent plane goes through the origin.

We know that the normal to the tangent plane is given by the grad(F) and can spilt the function into an upper and lower half to find that
$\displaystyle f(x,y,z)=\sqrt{x^2+y^2}\Rightarrow grad(F)=(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x ^2+y^2}},0)$. But when we evaluate at a point $\displaystyle P(x_0,y_0,z_0)$ we get:
$\displaystyle \frac{x_0}{\sqrt{x_0^2+y_0^2}}(x-x_0)+\frac{y_0}{\sqrt{x_0^2+y_0^2}}(y-y_0)=0$. From what we can see this need not be zero at $\displaystyle (0,0,0)$
• Jan 28th 2009, 09:44 PM
NonCommAlg
Quote:

Originally Posted by Pur

Given $\displaystyle z^2=x^2+y^2$, show that every tangent plane goes through the origin.

We know that the normal to the tangent plane is given by the grad(F) and can spilt the function into an upper and lower half to find that
$\displaystyle f(x,y,z)=\sqrt{x^2+y^2}\Rightarrow grad(F)=(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x ^2+y^2}},0)$. But when we evaluate at a point $\displaystyle P(x_0,y_0,z_0)$ we get:
$\displaystyle \frac{x_0}{\sqrt{x_0^2+y_0^2}}(x-x_0)+\frac{y_0}{\sqrt{x_0^2+y_0^2}}(y-y_0)=0$. From what we can see this need not be zero at $\displaystyle (0,0,0)$

to find the gradient you need to write your surface in the form $\displaystyle f(x,y,z)=0$ first. the equation of the cone is $\displaystyle f(x,y,z)=x^2+y^2-z^2=0.$ thus: $\displaystyle \nabla f(P)=(2x_0,2y_0,-2z_0),$ and so the

equation of the tangent plane at $\displaystyle P$ is: $\displaystyle 2x_0(x-x_0)+2y_0(y-y_0)-2z_0(z-z_0)=0,$ which after simplifying gives us: $\displaystyle x_0x+y_0y-z_0z=0. \ \Box$
• Jan 29th 2009, 10:02 AM
Pur
Thank you, why is it that we need to use IFT since $\displaystyle z=\sqrt{x^2+y^2}$ is a valid equation for the upper half?