回復 52# Ellipse 的帖子
填充 4. 坐標硬解法,注意稱性,將矩形的中心點取為原點,令\( A( - 1,\frac{1}{2}),C(1, - \frac{1}{2}),P(2t,t) \)
\( \vec{AP}\cdot \vec{CP} = 5{t^2} - \frac{5}{4} = \sqrt {5{t^2} + 3t + \frac{5}{4}} \cdot \sqrt {5{t^2} - 3t + \frac{5}{4}} \cdot \frac{{\sqrt 3 }}{{\sqrt {10} }}\)
\( \Rightarrow 25{(4{t^2} - 1)^2} = (20{t^2} + 12t + 5)(20{t^2} - 12t + 5) \cdot \frac{9}{{10}}\)
\( \Rightarrow 400{t^4} - 2504{t^2} + 25 = 0 \)
公式解可得 \( {t^2} = \frac{{2504 \pm 2596}}{{800}} = \frac{{25}}{4},\frac{1}{{100}} \Rightarrow t = \pm \frac{5}{2}, \pm \frac{1}{{10}} \)
其中 \( \pm \frac1{10} \) 檢驗應為 \( \cos \alpha = - \frac{\sqrt{3}}{\sqrt{10}} \)