Quadric Surfaces — Question 8

Question 8

An elliptic paraboloid has vertex $(1,-2,3)$, axis parallel to the $z$-axis, opens downward, and its trace in $z=-1$ is $$\frac{(x-1)^2}{8}+\frac{(y+2)^2}{4}=1.$$

Tasks

1. Reconstruct the surface equation.

2. Find the trace at $z=2$ in standard form.

3. State that trace’s semiaxes and verify the supplied trace.

Question 8 – Solution

Strategy Use translated form $z-3=-A(x-1)^2-B(y+2)^2$ and fit the known level.

Step 1: Use the known trace At $z=-1$, the vertical drop is $4$. The given ellipse is equivalent to $$\frac 12(x-1)^2+(y+2)^2=4.$$ Hence the surface equation is $$\boxed{z-3=-\frac 12(x-1)^2-(y+2)^2}.$$

Step 2: New trace Set $z=2$: $$1=\frac 12(x-1)^2+(y+2)^2,$$ or $$\boxed{\frac{(x-1)^2}{2}+(y+2)^2=1}.$$ Its semiaxes are $\sqrt 2$ and $1$. Substituting $z=-1$ back into the derived surface recovers the supplied ellipse.