Are there any point where the tangent plane to $f(x,y)=ye^{-x}$ is parallel to the plane $y+2z = 4$?

Question

I did try this question by finding the gradient and solving them but my answer was $(0,0,0)$. No way to check if its correct and I am unsure about it because it has couple of missing variables in either planes.

So, I need help. Thank you very much

Answer 1

With $z=f(x,y)$, the equation for the tangent plane is given by $z-f(x_0,y_0)=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$. By writing in the form $ax+by+cz+d=0$, it is clear that the normal direction is $(-f_x(x_0,y_0),-f_x(x_0,y_0),1)$.

The plane you have is $y+2z=4$, or $0x+y+2z-4=0$ so the normal direction is $(0,1,2)$. For two planes to be parallel, their normal directions must be the same. In other words, one must be a constant multiple of the other.

To match them, we want $(-2f_x(x_0,y_0),-2f_x(x_0,y_0),2)=(0,1,2)$. The problem is then solving two equations with two unknowns, $(x_0,y_0)$. For the first component, we want $-2f_x(x_0,y_0)=-2y_0e^{-x_0}=0$, which only occurs when $y_0=0$. For the second component, we want $-2f_y(x_0,y_0)=-2e^{-x_0}=1.$ This can never occur. Therefore, the tangent planei s never parallel to the plane $y+2z=4$.

Answer 2

Hint

The equation if the surface can be expressed as $$f(x,y,z)=z-ye^{-x}=0$$so the question is equivalent to

Is there any point on the surface whose gradient vector $\nabla f(x,y,z)$ is parallel to the perpendicular vector of the plane (which is $(0,1,2)$)?