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This problem feels really easy but I've been having a really hard time with it. I'm given an equation of a paraboloid $z=x^2+4y^2$ and told that an unknown plane, perpendicular to the $xy$ plane has a point $(2,1,8)$ in common with the paraboloid. The intersection between the plane and the paraboloid is a parabola with slope $0$ at the given point.

I'm told to find the equation of the plane. I've tried using the gradient vector but I found out that my approach is wrong. I tried to explicitly find the intersection between a plane with an equation $y=ax+d$ and the paraboloid equation, then differentiate it once to find out what $a$ and $d$ are so that the slope in $(2,1,8)$ is $0$, looking at the graphs in Mathematica it seems that I've got it wrong with both approaches. Looking for any suggestions on this, I'm really lost.

Edit: some information on the gradient approach. I calculated $\nabla z(x,y)=(2x,8y)$, then substituted $x$ and $y$ for $2$ and $1$ respectively. This should be perpendicular to the level curve $8=x^2+4y^2$, if I'm thinking correctly. Therefore, I can define a plane using $\nabla z(2,1)=(4,8)$ and using the fact that we know the plane is orthogonal to $xy$, therefore we use $(4,8,0)$ as the normal vector to the plane. So, by my chain of thought, the plane equation should be $4x+8y+d=0$, substituting $x$ and $y$ for $2$ and $1$ we get $d=-16$. Unless I messed up my Mathematica plot lots of times, this isn't right...

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  • $\begingroup$ I suggest you edit the question to give more detail (including equations) about your attempts to solve this. I think you could easily use the gradient to solve this, so it's unclear how you would have "found out that my approach is wrong" unless some error you made in the calculations caused a nonsensical result. $\endgroup$ – David K May 3 '17 at 21:46
  • $\begingroup$ I edited in some more information about how I used the gradient vector. $\endgroup$ – AstlyDichrar May 3 '17 at 22:00
  • $\begingroup$ Your attempt gave you the equation $4x+8y-16=0,$ which is equivalent to $4x+8y=16,$ which is equivalent to $x+2y=4,$ which is also shown in an answer below. So you already had the correct answer. $\endgroup$ – David K May 3 '17 at 22:07
  • $\begingroup$ Yeah, it looks like where I'm messing up is just graphing it on Mathematica. $\endgroup$ – AstlyDichrar May 3 '17 at 22:10
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If the slope of the parabolic intersection at the point $(2,1,8)$ is zero, then that point is the vertex of the parabola. So the plane being sought will be tangent to the level curve $x^2+4y^2=8$ at the point $(2,1,0)$. That line has slope $m=-\dfrac{1}{2}$.

So the intersection of the plane being sought and the $xy$-plane will be the line with slope $m=-\dfrac{1}{2}$ and containing $(2,1,0)$, namely $y=-\dfrac{1}{2}x+2$ which is also the equation of the plane. Tangent to ellipse at point

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  • $\begingroup$ I don't think this is correct: if you draw the level curves, you will see that the line from the origin to $(2,1)$ is not perpendicular to the plane. The condition should be that the intersection of the plane with the $xy$ plane is the tangent to the level curve at point $(2,1)$. $\endgroup$ – NickD May 3 '17 at 22:20
  • $\begingroup$ @Nick You are correct. I am amending my answer. $\endgroup$ – John Wayland Bales May 4 '17 at 2:43
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    $\begingroup$ This is much easier to understand than the other methods, thanks! $\endgroup$ – AstlyDichrar May 5 '17 at 0:57
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Your approach is correct. The plane is $y = ax + d$, so the intersection is a parabola $z = x^2 + 4(ax + d)^2$. The point $(2,1,8)$ gives two restrictions on $a$ and $d$: $1 = 2a + d$ (because point lies on the plane) and $8 = 2^2 + 4(2a + d)^2$ (because point lies on a paraboloid). Notice that they are identical, and conclude that $d = 1-2a$. The parabola equation becomes

$$z = x^2 + 4(ax + 1 - 2a)^2$$

Differentiate wrt $x$. The derivative must vanish at $x = 2$. This leads to a quadratic equation for $a$. Solve it.

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  • $\begingroup$ It appears both ways give the same result, I must be messing up my Mathematica plot... $\endgroup$ – AstlyDichrar May 3 '17 at 22:04
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The normal to the paraboloid and $(2,1,8) = (4,8,-1)$

$4x + 8y - z = 8$ is a plane tangent to the paraboliod at the point (1,2,8)

I say that our target plane intersects the plane described above and forms a line that is parallel to the $xy$ plane. This way the parabola formed by the intersection of our target plane and the paraboliod can be tangent to this line and at a minimum.

$x + 2y = 4$

lets check.

substitute $x = 4 - 2y$

and hopefully we get a minimum at (1,8)

$z = (4-2y)^2 + 4y^2\\ z = 8 y^2 - 16 y + 16$

looks good.

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