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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

  1. 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;

    2xi-j+2zk=7i+2j+3k
    becomes
    2xi-j+2xk=(-7/2)i-j+(-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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  2. normal vector to f(x,y,z)=0 is

    so, for the paraboloid,
    n = fxi + fyj + fzk
    = 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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