Re: [中學] 103新店高三段考橢圓參數式

看板Math作者kerwinhui (kezza)時間8年前 (2015/09/10 00:14)推噓3(3推 0噓 3→)

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※ 引述《dramatic0306 (悶騷)》之銘言： : 題目是這樣 : 圓(x+4)^2 +(y-3)^2 = 1上一點P : 橢圓(x^2)/9 + (y^2)/1 = 1上一點Q : 求PQ線段最大值? : 我知道利用圓心到橢圓求極值再加圓半徑,可是問題會卡在圓心到橢圓利用參數式,代兩點 : 間距離公式後會配不出來（不知道怎麼配） : 感謝大家撥空看看了~~!! 的確不好配，會要解一元四次方程橢圓點 Q(3 cos Θ, sin Θ) 距離 (-4,3) 極值，所以 (-4-3cosΘ,3-sinΘ) . (3 sinΘ,-cosΘ) = 0 即 12sinΘ+8sinΘcosΘ+3cosΘ=0 設 t=tan(Θ/2), 則 sinΘ=2t/(1+t^2), cosΘ=(1-t^2)/(1+t^2) 可得 3 t^4 - 8 t^3 - 40 t - 3 = 0 有兩個實根，約在 t=-0.075 和 t=3.675 如果你想看，代進 http://www.curtisbright.com/quartic/quartic-png.html 的兩個實根是： t = 2/3+1/(3 sqrt(2/(8-(89 3^(2/3))/(2592+sqrt(8833371))^(1/3)+(3 (2592+sqrt(8833371)))^(1/3))))-1/2 sqrt(32/9-(2 (2592+sqrt(8833371))^(1/3))/(3 3^(2/3))+178/(3 (3 (2592+sqrt(8833371)))^(1/3))+424/9 sqrt(2/(8-(89 3^(2/3))/(2592+sqrt(8833371))^(1/3)+(3 (2592+sqrt(8833371)))^(1/3)))) 和 t = 2/3+1/(3 sqrt(2/(8-(89 3^(2/3))/(2592+sqrt(8833371))^(1/3)+(3 (2592+sqrt(8833371)))^(1/3))))+1/2 sqrt(32/9-(2 (2592+sqrt(8833371))^(1/3))/(3 3^(2/3))+178/(3 (3 (2592+sqrt(8833371)))^(1/3))+424/9 sqrt(2/(8-(89 3^(2/3))/(2592+sqrt(8833371))^(1/3)+(3 (2592+sqrt(8833371)))^(1/3)))) -- 『我思故我在』怎樣從法文變成拉丁文的： je pense, donc je suis --- René Descartes, Discours de la Méthode (1637) ego sum, ego existo --- ____, Meditationes de Prima Philosophia (1641) ego cogito, ergo sum --- ____, Principia Philosophiae (1644) -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 111.235.202.95 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1441815280.A.33A.html

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