Question

At what point on the paraboloid y = x^2 +z^2 is the tangent plane parallel to the plane 3x+2y+3z = 8?

Answer 1

At what point on the paraboloid \[y = x^2 +z^2\] is the tangent plane parallel to the plane \[3x+2y+3z = 8\]?

So, finding normal vector of the tangent plane, you would subtract y to the other side to get \[0=x^2-y+z^2\] which gives you an equation of the form k = f(x, y, z) where k is a constant (in this case 0) so the normal of the tangent plane should then be the gradient vector \[<2x, -1, 2z>\] Now, my understanding of determining if two planes are parallel is if there normal vectors are scalar multiples of each other, in which case take the two normal vectors: \[<2x,-1, 2z>\] \[<3,2,3>\] and set the parts equal to each other like \[2x = 3c\]\[-1 = 2c\]\[2z = 3c\] where c is a constant.
Solve for c using the y-components so you get c = -0.5, then solve for x and z using c = -0.5 and you should get the point \[(\frac{-3}{4}, 1, \frac{-3}{4})\] right? Except that's wrong so not right I guess... :/

Answer 2

exxcellent! :D

Answer 3

Oh, not quite. I need to plug -3/4 into x and z in the original equation to get (-3/4, 9/8, -3/4) ...