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📅 May 18, 2026 👤 scraface

May 18, 2026

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Now substitute $
ho = \sqrt{x^2 + y^2 + z^2} $:
2\sqrt{x^2 + y^2 + z^2} = 4 - z.
Now square both sides:
4(x^2 + y^2 + z^2) = (4 - z)^2 = 16 - 8z + z^2.
4x^2 + 4y^2 + 4z^2 = 16 - 8z + z^2 \Rightarrow 4x^2 + 4y^2 + 3z^2 + 8z...
Divide by 4 to simplify:
x^2 + y^2 + rac{3}{4}z^2 + 2z - 4 = 0.
Now complete the square for $ z $:
rac{3}{4}z^2 + 2z = rac{3}{4}(z^2 + rac{8}{3}z) = rac{3}{4}\left( \...

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