Finding normal and tangent plane where one of the derivatives = 1?

Question

I am supposed to find the tangent and normal planes for $z=2x^2+y^2$ at $M = (1,-1,3)$. But my issue is that when I find the derivative of $z$, I get $1$. How am I supposed to incorporate $z = 3$ (from the point $M$) into the tangent plane equation?

Answer 1

$$ f(x,y,z)=2x^2+y^2-z=0 $$ The normal vector is proportional to the gradient of $f$

$$ \vec \nabla f(x,y,z) = (4x, 2y, -1) $$ at $(-1,1,3)$ this has the value $$ \vec \nabla f(-1,1,3) = (-4, 2, -1) $$ So the equation of the tangent plane is

$$-4x+2y-z =-4(-1)+2(1)-(3)=3 $$

and the parametric equation of the normal is

$$(x(t) ,y(t) ,z(t) ) = (-1,1,3)+(-4,2,-1)t $$