回復 11# Harris 的帖子
設\(\displaystyle A(s,t),B(\frac{t}{z},-\frac{s}{z})\)
則\(\cases{\displaystyle \frac{s^2}{25}+\frac{t^2}{16}=1\ldots(1)\cr
\frac{t^2}{25}+\frac{s^2}{16}=z^2\ldots(2)}\)
\(\displaystyle (1)+(2) \Rightarrow (s^2+t^2)(\frac{1}{25}+\frac{1}{16})=1+z^2\)
所求\(\displaystyle =\frac{1+z^2}{s^2+t^2}=\frac{1}{25}+\frac{1}{16}=\frac{41}{400}\)
用此法可一目了然定值的原因.
