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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.

Wilson Brians
  • 747

3 Answers3

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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 $(2,-1,4), (-2,1,-4)$.

DeepSea
  • 77,651
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You have computed $$\nabla F=(2xy + y^2 + 3, x^{2} + 2yx, -1).$$ Now, $(\nabla F)(x,y,z)$ is perpendicular to the tangent plane at $(x,y,z).$ So, if the tangent plane is parallel to $OXY$ then the gradient must be parallel to $(0,0,1)$ (a vector normal to the plane $OXY$).

Thus, you need to solve the system

$$\begin{cases}2xy+y^2+3&=0,\\ x^2+2xy&=0.\end{cases}$$ From the seconde equation it is $2xy=-x^2.$ Substituting in the first equation you have $$x^2-y^2=3.$$ This is the equation of a hyperbole. At any point of this hyperbole the tangent plane of the surface is parallel to $OXY.$

mfl
  • 29,399
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Hint

A plane parallel to $xy-$plane will have an equation of the form

$$z=z_0+0.x+0.y.$$

It's your turn to solve the system

$$\frac{\partial F}{\partial x}(x_0,y_0)=0$$

$$\frac{\partial F}{\partial y}(x_0,y_0)=0$$

hamam_Abdallah
  • 62,951