Quadric Surfaces — Question 6

Question 6 – Solution

Strategy Substitute each fixed coordinate and classify the resulting planar curve.

Vertical traces In $y=0$, $z=x^2$ opens upward. In $x=0$, $z=-4y^2$ opens downward. Their opposite concavities create the saddle.

Horizontal traces At $z=0$, $x^2-4y^2=0$, so $\boxed{x=\pm 2y}$, two intersecting lines. For $k>0$, $x^2-4y^2=k$ is a hyperbola opening in the $x$-direction; for $k<0$, it opens in the $y$-direction.

Conclusion The origin is neither a local maximum nor minimum because nearby traces rise in one direction and fall in the perpendicular direction.