At what pts. on the surface $z = x^{2}y + y^{2}x + 3x$ is the tangent plane parallel to the $xy$-plane?
So first I define a function $F(x, y, z) = x^{2}y + y^{2}x + 3x – z$ which has gradient $grad F = (2xy + y^2 + 3, x^{2} + 2yx, -1)$. So the equation of our tangent plane is: $(2xy + y^{2} + 3)(x – x_{0}) + (x^{2} + 2yx)(y – y_{0}) – (z – z_{0}) = 0$.
So we get as a normal line:
$r(t) = (x_{0} + (2x_{0}y_{0} + y_{0}^{2} + 3)t, y_{0} + (x_{0}^{2} +2y_{0}x_{0})t, z_{0} – t)$.
Now I know two planes are parallel if their normal lines are parallel, but I'm not quite sure how to complete the problem.
Best Answer
If the tangent plane to the surface defined by $f(x,y,z) = 0$ with $f(x,y,z) = x^2y+y^2x+3x-z$ at the point $P = (x,y,z)$ is parallel to the $xy$ plane, then its normal line must be parallel to the vector $\vec{a} = <0,0,1>$. But $\vec{n} = \nabla{f}= <2xy+y^2+3, x^2+2xy, -1>$. Thus we must have: $2xy+y^2+3 = 0 = x^2+2xy$. Observe that $x \neq 0$ since $y^2+3 > 0$. Thus: $x(x+2y) = 0 \implies x = -2y\implies -4y^2+y^2+3 = 0\implies y^2 = 1 \implies y = \pm 1\implies x = \mp 2$. Thus the points are $(2,-1,4), (-2,1,-4)$.