# Find the equation of the tangent plane to the cone x^2+y^2=Z^2 at the point (x0; y0; z0)`!=(0,0,0)` using the gradient method. Note: The equation of the cone must be in level surface form. Show...

Find the equation of the tangent plane to the cone x^2+y^2=Z^2 at the point (x0; y0; z0)`!=(0,0,0)`

using the gradient method. Note: The equation of the cone must be in level surface form. Show that this tangent plane passes through the origin.

### 1 Answer | Add Yours

Given

`z^2=x^2+y^2` (i)

`z^2_0=x^2_0+y^2_0`

`z_0=+-sqrt(x^2_0+y^2_0)`

Differentiate (i) with respect to x and y partially

`2z(delz)/(delx)=2x`

`(delz)/(delx)=x/z`

`((delz)/(delx))_{(x_0,y_0,z_0)}=x_0/z_0`

`=(+-x_0)/sqrt(x^2_0+y^2_0)`

`2z(delz)/(dely)=2y`

`(delz)/(dely)=y/z`

`((delz)/(dely))_{(x_0,y_0,z_0)}=y_0/z_0=(+-y_0)/sqrt(x^_0+y^2_0)`

Thus equation of the tangent plane at `(x_0,y_0,z_0)` is

`z-z_0=((+-x_0)/sqrt(x^2_0+y^2_0))(x-x_0)+((+-y_0)/(sqrt(x^2_0+y^2_0)))(y-y_0)`

`(z-z_0)(sqrt(x^2_0+y^2_0))=+-x_0(x-x_0)+-y_0(y-y_0)`

`` `(z-z_0)sqrt(x^2_0+y^2_0)=+-(x_0x-x^2_0+y_0y-y^2_0)`

`z=z_0+-(x_0x+y_0y-x^2_0-y^2_0)/sqrt(x^2_0+y^2_0)`

`x_(mn)`