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📅 May 3, 2026 👤 scraface

May 03, 2026

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But the orbit points are only on the ellipse — so only the points wher...
Since only \( (0,8) \) and \( (0,-8) \) satisfy both, there are **2**...
But earlier algebra confirms only two solutions.
Therefore, the signal reaches **2 points**.
Wait — contradiction.
Let’s re-solve the system:
\frac{x^2}{100} + \frac{y^2}{64} = 1 \quad \text{(ellipse)}
x^2 + y^2 = 64 \quad \text{(circle)}
From second, \( y^2 = 64 - x^2 \). Substitute:
x^2 \left( \frac{64 - 100}{6400} \right) = 0 \Rightarrow x^2 \cdot (-3...

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