[線代] Quadratic curve

看板Math作者 (憶)時間12年前 (2011/09/26 21:58), 編輯推噓0(005)
留言5則, 2人參與, 最新討論串1/1
In |R^3, let C be a circle lying on a plane P which does not pass through the origin (0,0,0). Let Γ={(x,y)∈|R^2|(tx,ty,t)∈C for some t∈|R} Show that Γ is a quadratic curve. Namely, there is a polynomail F(X,Y) of total degree 2 such that Γ={(x,y)∈|R^2|F(x,y)=0} Is Γ is ellipse, a parabola or a hyperbola? 感謝大家給與想法 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.240.208.100

09/27 05:46, , 1F
O到C 拉出來的圓錐 與平面 Z=1 切出來的曲線就是Γ
09/27 05:46, 1F

09/27 21:35, , 2F
感謝S大的解釋 可以理解
09/27 21:35, 2F

09/27 21:35, , 3F
不過要怎麼化成代數式子呢? 因為那個圓錐似乎無法判
09/27 21:35, 3F

09/27 21:36, , 4F
別長甚麼樣子(畢竟C和P都是會變動的)
09/27 21:36, 4F

09/27 21:37, , 5F
雖然猜測起來像是橢圓?
09/27 21:37, 5F
文章代碼(AID): #1EW8Jo6a (Math)