# Find the point at which the tangent plane is parallel to a given plane

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.

• Perhaps I misunderstand the question, but finding a point (on a plane) "at which the plane is parallel to another plane"...? That sounds strange since planes are either parallel or not (it does not "occur" at a certain point alone). Aug 11, 2021 at 14:54
• Two planes are parallels if the both have the same normal vector. To find the plane you need a normal vector and one point. Aug 11, 2021 at 14:54
• Ah hold on, I've mis-read the question. I'll edit it now and it should make sense, dear lord I need glasses Aug 11, 2021 at 14:56
• Yeah you're right, I can't use the tangent plane I've already found because that is the tangent plane only at the point $(-1,1,2)$, therefore I obviously can't find what I'm looking for Aug 11, 2021 at 14:59
• I'm still confused how they found the answer so quickly and with no working out Aug 11, 2021 at 15:00

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 $$(2)$$)
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.
• Ohhhh, that makes sense, after correctly reading the question I managed to figure out $x$ and $y$ but $z$ being $5$ confused me, I get it all now, thanks a lot for the quick replies and top quality explanations Aug 11, 2021 at 15:12