This is post is a continuation of a previous post I made: Find the equation of the intersection of two tangent planes this is the next part of that question but is very loosely related so I thought I should make a new post.
I have the parabaloid $(1):$ $z=x^2 +y^2$
I'm asked to find the point on this parabaloid where its tangent plane is parallel to the plane:$(2):$ $4x+8y-2z=10$
What I've set up is this: I need to find a point where the vector $(-2x,-2y,1)$ (obtained by finding the gradient of my parabaloid $(1)$) is a parallel to the vector $(4,8,-2)$ (obtained by finding the gradient of plane $(2)$)
On the answer sheet I have available to me, there is no working out shown and it simply says "the point this occurs at is $(1,2,5)$. The fact that they show no working out leads me to believe there is a quick/simple way to find this point, similar to how a simple cross product gave me my solution in my previous post.
If anyone could guide me in the right direction or show me what I should do it would really help.
Thanks in advance.
Alright, the question makes sense now.
So you want $(-2x,-2y,1)$ to be parallel to $(4,8,-2)$ which means they should be scalar multiples, so you're looking for some $k \in \mathbb{R}$ such that: $$(4,8,-2)=k(-2x,-2y,1)$$ This comes down to a simple system of equations but solving it is easily done by inspection since for $z$ you immediately have $k=-2$ and then $x=1$ and $y=2$ follow quickly, $z$ follows from the equation of the paraboloid.