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One focus of the ellipse \(\frac{x^2}{2} + y^2 = 1\) is at \(F = (1,0).\) There exists a point \(P = (p,0),\) where \(p > 0,\) such that for any chord \(\overline{AB}\) that passes through \(F,\) angles \(\angle APF\) and \(\angle BPF\) are equal. Find \(p.\)

 Oct 11, 2020
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p = 4.

 Oct 11, 2020

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