At what point on the paraboloid
y = x2 + z2
is the tangent plane parallel to the plane
7x + 2y + 3z = 2?
(If an answer does not exist, enter DNE.)
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2 answers

In other words,
divide the plane's normal vector so that the y components match (INCLUDING SIGN), then solve for the two remaining variables.
For the example above this would mean;
2xij+2zk=7i+2j+3k
becomes
2xij+2xk=(7/2)ij+(3/2)k
Now solve for x and z like so:
2x=(7/2) => x=(7/2)/2 = 7/4
2z=(3/2) => z=(3/2)/2 = 3/4
Now for y, just use the original equation (y=x^2+z^2), since we now know x and z, simply plug them in to get y like so:
y= (7/4)^2+(3/4)^2 = 29/8
So your coordinates are:
(7/4, 29/8, 3/4)
Happy trails. 👍
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answered by Josh 
normal vector to f(x,y,z)=0 is
so, for the paraboloid,
n = f_{x}i + f_{y}j + f_{z}k
= 2xi  j + 2zk
for the plane,
n = 7i + 2j + 3k
Now manipulate the directions and magnitudes so the two normals are the same. 👍
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answered by Steve
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