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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.

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  • $\begingroup$ 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). $\endgroup$
    – StackTD
    Aug 11, 2021 at 14:54
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    $\begingroup$ Two planes are parallels if the both have the same normal vector. To find the plane you need a normal vector and one point. $\endgroup$ Aug 11, 2021 at 14:54
  • $\begingroup$ Ah hold on, I've mis-read the question. I'll edit it now and it should make sense, dear lord I need glasses $\endgroup$
    – Charlie P
    Aug 11, 2021 at 14:56
  • $\begingroup$ 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 $\endgroup$
    – Charlie P
    Aug 11, 2021 at 14:59
  • $\begingroup$ I'm still confused how they found the answer so quickly and with no working out $\endgroup$
    – Charlie P
    Aug 11, 2021 at 15:00

1 Answer 1

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Alright, the question makes sense now.

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.

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    $\begingroup$ 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 $\endgroup$
    – Charlie P
    Aug 11, 2021 at 15:12

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