<em>I hope this helps.</em>
<em>Answer:-</em>
<em> ∴x−3y−4z=0</em>
<em />
<em>Explanation:</em>
<em />
<em>First we rearrange the equation of the surface into the form f(x,y,z)=0</em>
<em />
<em> x2+2z2=y2</em>
<em> ∴x2−y2+2z2=0</em>
<em />
<em>And so we have our function:</em>
<em />
<em> f(x,y,z)=x2−y2+2z2</em>
<em />
<em>In order to find the normal at any particular point in vector space we use the Del, or gradient operator:</em>
<em />
<em> ∇f(x,y,z)=∂f∂xˆi+∂f∂yˆj+∂f∂zˆk</em>
<em />
<em>remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:</em>
<em />
<em> ∇f=(∂∂x(x2−y2+2z2))ˆi+</em>
<em> (∂∂y(x2−y2+2z2))ˆj+</em>
<em> (∂∂z(x2−y2+2z2))ˆk</em>
<em> =2xˆi−2yˆj+4zˆk</em>
<em />
<em>So for the particular point (1,3,−2) the normal vector to the surface is given by:</em>
<em />
<em> ∇f(1,3,−2)=2ˆi−6ˆj−8ˆk</em>
<em />
<em>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:</em>
<em />
<em> →r⋅→n=→a⋅→n</em>
<em />
<em>Where →r=⎛⎜⎝xyz⎞⎟⎠; →n=⎛⎜⎝2−6−8⎞⎟⎠, is the normal vector and a is any point in the plane</em>
<em />
<em>Hence, the tangent plane equation is:</em>
<em />
<em> ⎛⎜⎝xyz⎞⎟⎠⋅⎛⎜⎝2−6−8⎞⎟⎠=⎛⎜⎝13−2⎞⎟⎠⋅⎛⎜⎝2−6−8⎞⎟⎠</em>
<em> ∴(x)(2)+(y)(−6)+(z)(−2)=(1)(2)+(3)(−6)+(−2)(−8)</em>
<em> ∴2x−6y−8z=2−18+16</em>
<em> ∴2x−6y−8z=0</em>
<em> ∴x−3y−4z=0</em>
