# Invalid method of finding points on a surface making tangent plane parallel to a plane?

What are the points on the surface $$z = x^2+4y^2$$ in which the tangent plane is parallel to the plane $$x+y+z=0$$?

I tried solving this problem today but my teacher told me my method was invalid, although I got the same answer. My approach was to use the standard ellipse equation and getting the center coordinates for a solution. I rewrote the equations and made them equal: $$x^2+4y^2=-x-y$$. By solving this, I got the solution $$(x, y) = (\frac{-1}{2}, \frac{-1}{8})$$ (also centre coordinates) to the equation which makes $$z=\frac{5}{16}$$ which is the correct answer according to our book.

So, can anyone please explain to me why my method is invalid? Is it maybe because in other cases it doesn't include all of the possible points?

• What do you mean by points parallel to a plane? Commented Jan 20, 2023 at 20:35
• Also, I don't understand why (or how) you got $(x,y)=(-1/2,-1/8)$. $(x,y)=(0,0)$ is also a solution. And there are infinitely many other. Commented Jan 20, 2023 at 20:38
• I tried creating an ellipse by rewriting the equations to use the ellipse standard equation and got the center coordinates (-1/2, -1/8). Commented Jan 20, 2023 at 20:46
• It is possible to show that you get the right answer with this method for quadratic polynomials. But there is no guarantee for any other surfaces. Commented Jan 20, 2023 at 22:01

If the tangent plane to $$z = x^2+4y^2$$ is parallel to the plane $$x + y + z = 0$$ the tangent plane is of the form $$x + y + z = c$$ where $$c \in \mathbb R$$. Now if you can find $$c$$ such that a plane $$x + y + z = c$$ intersects $$z = x^2+4y^2$$ in a single points then you are done. Looking at the equation $$x^2+4y^2 + x + y = c$$ you only find one solution when the ellipse is reduced to a single point : the center of the ellipse. This center is also the center of the ellipse
$$x^2+4y^2 +x + y = 0$$