Actually, write in standard form:
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- Divide both sides by $ rac{16}{3} $:
- rac{x^2}{ rac{64}{9}} + rac{y^2}{ rac{64}{3}} + rac{\left(z + rac{4}{3}
- ight)^2}{ rac{64}{9}} = 1 \quad ext{(after scaling)}.
- ight)^2}{ rac{64}{9}} = 1.
- Note that $ rac{64}{3} = rac{192}{9} > rac{64}{9} $, so the $ z $-axis term has larger denominator â but $ x^2 $ and $ \left(z + rac{4}{3}
- ight)^2 $ term has same denominator $ rac{64}{9} $, and $ y^2 $ has $ rac{192}{9} $, so it's **elliptical** in $ x/z $-plane. But more precisely, since the equation resembles $ rac{x^2 + y^2}{a^2} + rac{(z - z_0)^2}{b^2} = 1 $, with $ a
