Quadric Surfaces — Question 2

Question 2 – Solution

Strategy Read the signs and inspect traces one coordinate at a time.

Step 1: Classification Two positive squared terms and one negative term equal $1$, so this is a hyperboloid of one sheet. The negative term is the $z$-term, hence the axis is the $z$-axis.

Step 2: Horizontal traces Setting $z=c$ gives $$\frac{x^2}{9}+\frac{y^2}{4}=1+\frac{c^2}{16}.$$ Every right side is positive, so every horizontal plane cuts an ellipse. At $c=0$ the waist ellipse has semiaxes $3$ and $2$.

Step 3: Vertical traces At $y=0$, $$\frac{x^2}{9}-\frac{z^2}{16}=1,$$ and at $x=0$, $$\frac{y^2}{4}-\frac{z^2}{16}=1.$$ Both are hyperbolas opening perpendicular to the $z$-axis.

Conclusion Because a nonempty ellipse exists for every real $z=c$ and varies continuously with $c$, the surface is connected.