Question

Find x,y,z : 4x^2+2y^2+2z^2-4xy-4xz+2yz-2y+6z+10=0

Answer 1

I hope this helps.

Answer:-

   ∴x−3y−4z=0

Explanation:

First we rearrange the equation of the surface into the form f(x,y,z)=0

   x2+2z2=y2

   ∴x2−y2+2z2=0

And so we have our function:

   f(x,y,z)=x2−y2+2z2

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

   ∇f(x,y,z)=∂f∂xˆi+∂f∂yˆj+∂f∂zˆk

remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

   ∇f=(∂∂x(x2−y2+2z2))ˆi+

                    (∂∂y(x2−y2+2z2))ˆj+

                    (∂∂z(x2−y2+2z2))ˆk

      =2xˆi−2yˆj+4zˆk

So for the particular point (1,3,−2) the normal vector to the surface is given by:

   ∇f(1,3,−2)=2ˆi−6ˆj−8ˆk

So the tangent plane to the surface x2+2z2=y2 has this normal vector and it also passes though the point (1,3,−2). It will therefore have a vector equation of the form:

   →r⋅→n=→a⋅→n

Where →r=⎛⎜⎝xyz⎞⎟⎠; →n=⎛⎜⎝2−6−8⎞⎟⎠, is the normal vector and a is any point in the plane

Hence, the tangent plane equation is:

   ⎛⎜⎝xyz⎞⎟⎠⋅⎛⎜⎝2−6−8⎞⎟⎠=⎛⎜⎝13−2⎞⎟⎠⋅⎛⎜⎝2−6−8⎞⎟⎠

   ∴(x)(2)+(y)(−6)+(z)(−2)=(1)(2)+(3)(−6)+(−2)(−8)

   ∴2x−6y−8z=2−18+16

   ∴2x−6y−8z=0

   ∴x−3y−4z=0