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
