lgzrelv的部落格

標題:

F5 Maths circle tangents

發問:

Two tangents are drawn to the circle x^2 + y^2 - 6x + 6y - 2 = 0. If the two tangents are parallel to the straight line L: 2x+ y + 1 = 0, finda) the equations of the two tangents b) the coordinates of the point of contace for each tangent found in (a) 更新: Could you give me more detailed steps from (4c + 18)^2 - 20(c^2 + 6c - 2) = 0 to c^2 - 6c - 91 = 0 更新 2: From x^2 + 2x + 1 = 0 to x = -1 then y = -5. So, contact point (-1, -5) How to know that x = -1 then y = -5 ? 更新 3: How to know x = -1 ? 我可能睇miss左D野, 希望您耐心指導^^ 更新 4: http://hk.knowledge.yahoo.com/question/question?qid=7011012401441 唔該答埋呢到, 我會比哂最佳您

最佳解答:

(a) The slope of the tangents = -2 Let the equation of the tangent is y = -2x + c, Sub. into C: x^2 + y^2 - 6x + 6y - 2 = 0 x^2 + (c - 2x)^2 - 6x + 6(c - 2x) - 2 = 0 5x^2 - (4c + 18)x + (c^2 + 6c - 2) = 0 Discriminant = 0 (4c + 18)^2 - 20(c^2 + 6c - 2) = 0 c^2 - 6c - 91 = 0 (c - 13)(c + 7) = 0 c = -7 or c = 13 The equations of the tangents are y = -2x - 7 and y = -2x + 13 (b) Consider 5x^2 - (4c + 18)x + (c^2 + 6c - 2) = 0 when c = -7 x^2 + 2x + 1 = 0 x = -1 then y = -5. So, contact point (-1, -5) when c = 13 x^2 - 14x + 49 = 0 x = 7 then y = -1. So, contact point (7, -1) 2011-01-26 19:21:41 補充： (i) (4c + 18)^2 - 20(c^2 + 6c - 2) = 0 16c^2 + 144c + 324 - 20c^2 - 120c + 40 = 0 -4c^2 + 24c + 364 = 0 c^2 - 6c - 91 = 0 (ii) y = -2x - 7

其他解答:

• Canon G11 vs Nikon L110@1@
• 有沒有真正的婚姻服務所呢._0
• 請問元朗搭咩車去伊利沙伯醫院最快？
• university of sunderland定sheffield hallam讀design好d-@1@
• 京都遊(嵐山去半天)
• Differentiation 12
• 噴漆天拿水
• 證明質數方法@1@
• 請問深圳有冇一些唱k的地方好介紹---
• 象棋入門,要求教人捉棋的網站或software

 

此文章來自奇摩知識+如有不便請留言告知

0D7DAC4E7B8CAAC5