Quadric Surfaces — Question 5

Question 5 – Solution

Strategy Rearrange the trace family and interpret both its signs and its existence condition.

Step 1: Reconstruct Move terms and divide by $4$: $$x^2+y^2=\frac 49z^2-4\Longrightarrow \boxed{\frac{z^2}{9}-\frac{x^2}{4}-\frac{y^2}{4}=1}.$$

Step 2: Geometry One positive square and two negative squares equal $1$, so this is a two-sheet hyperboloid with $z$-axis. Setting $x=y=0$ gives vertices $\boxed{(0,0,\pm 3)}$.

Step 3: Vertical traces Setting $y=0$ and $x=0$ gives $$\boxed{\frac{z^2}{9}-\frac{x^2}{4}=1},\qquad \boxed{\frac{z^2}{9}-\frac{y^2}{4}=1}.$$

Step 4: Gap Since $x^2+y^2\ge 0$, the given trace requires $z^2/9-1\ge 0$, hence $|z|\ge 3$. No surface points lie in the central slab $-3<z<3$.