Tangent plane parallel to given plane

Question

Let $S$ be the surface in $R^3$ given by $z = \sqrt{2x^2+y^4+1}$ and $P$ be the plane given by $x-y-z=0$. Find an equation of the plane tangent to $S$ and parallel to $P$.

Progress I have made so far: In order for gradient to be parallel to normal vector of the plane $P$, $y^2$ should be equal to $1$. But the question is, how can I finish the problem which asks me the equation of the plane? Any help will be strongly appreciated!

Answer 1

Let $h(x,y,z)=2x^2+y^4+1-z^2$ and let $S^\star=\{(x,y,z)\in\Bbb R^3\mid h(x,y,z)=0\}$; then $S=\{(x,y,z)\in S^\star\mid z>0\}$. You have $\nabla h(x,y,z)=(4x,4y^4,-2z)$ and you want to find a point of $S$ such that $\nabla h(x,y,z)$ is a multiple of $(1,-1,-1)$. So, you solve the system:$$\left\{\begin{array}{l}4x=\lambda\\4y^3=-\lambda\\-2z=-\lambda\\h(x,y,z)=0.\end{array}\right.$$It is not hard to see that the only solution is $x=1$, $y=-1$, $z=2$, and $\lambda=4$. So, the plane that you're after is the plane passing through $(1,-1,2)$ which is parallel to the plane $x-y-z=0$, which is the plane $x-y-z=0$ itself.

Answer 2

You should be able to further determine that $y=-1$. Apply the similar analysis to $x$ and $z$ to obtain $x=1$ and $z=2$. Given this tangential point $(1,-1,2)$ and the normal vector $(1,-1,-1)$, the equation of the plane is

$$(x-1,y+1,z-2)\cdot(1,-1,-1)=0$$