Quadric Surfaces — Question 7

Question 7 – Solution

Strategy Substitute the fixed height and rewrite the remaining equation in standard ellipse form.

Step 1: Substitute With $z=1$, $$\frac{x^2}{16}+\frac{y^2}{9}+\frac 14=1\Longrightarrow\frac{x^2}{16}+\frac{y^2}{9}=\frac 34.$$ Divide by $3/4$: $$\boxed{\frac{x^2}{12}+\frac{y^2}{27/4}=1}.$$ The semiaxes are $\boxed{2\sqrt 3}$ and $\boxed{3\sqrt 3/2}$.

Step 2: Area Therefore $$A=\pi ab=\pi(2\sqrt 3)(3\sqrt 3/2)=\boxed{9\pi}.$$

Comparison At $z=0$, the semiaxes are $4$ and $3$, so the central area is $12\pi$. The section at $z=1$ has area $3/4$ as large, matching the right side $3/4$ before normalization.