Quadric Surfaces — Question 2

Question 2 – Solution

Strategy Complete squares and inspect the signs. Then substitute fixed $z$-levels.

Standard form Simplification gives $$(x-1)^2+4(y+2)^2-(z-2)^2=25, \quad\boxed{\frac{(x-1)^2}{25}+\frac{(y+2)^2}{25/4}-\frac{(z-2)^2}{25}=1}.$$ It is a hyperboloid of one sheet centered at $(1,-2,2)$ with axis parallel to $z$.

Traces At $z=2$ the waist ellipse is $(x-1)^2+4(y+2)^2=25$. At $z=2+k$ it becomes $(x-1)^2+4(y+2)^2=25+k^2$, so ellipses expand as $|k|$ grows.