Actually, write in standard form:

February 22, 2026

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

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