I'm asked to find the distance to the origin of the tangent plane of the surface $x^2-y^2+2z^2=5$ in the point $(2,-1,1)$. This seems to be an optimization problem with constraint, which I think can be solved with Lagrange multiplier.
The tangent plane is of the form $f_x(p_0)(x-x_0)+f_y(p_0)(y-y_0)+f_z(p_0)(z-z_0)$ where $f_x=2x, f_y=-2y, f_z=4z$. So in that point, we get $4(x-2)+2(y+1)+4(z-1)$. This will be our constraint $g(x,y,z)$
The function we want to optimize is the Euclidian distance squared $(x-0)^2+(y-0)^2+(z-0)^2:=f(x,y,z)$
So with lagrange :$(2x,2y,2z)=\lambda(4,2,4)$ so $(x,y,z)=\lambda(2,1,2)$.
So $\lambda=\frac{x}{2}=y=\frac{z}{2}$
So $x=z=2y$
My questions are :
1) Is what I did until now correct ?
2) If yes, how am I supposed to find the given point(s) now ? I have everything expressed in term of one variable but how do I find the value of this variable ? I thought about plugging $x$ and $z$ in terms of $y$ into the original surface equation $x^2-y^2+2z^2=5$ but we're asked to find the distance of the tangent plane of that equation at a point, so the tangent plane of that point isn't necessarily member of that surface, or am I wrong ?
Thanks for your help !
Hint
The tangent plane is perpendicular to $\nabla f$ where $f(x,y,z)=x^2-y^2+2z^2.$ (Why?) So its equation is given by $$2x-2y+4z=D.$$
Moreover, the plane contains the point $(2,-1,1).$ (Why?)
So, you have to get the distance from the origin to a plane.