MQ21 --- Coordinate Geometry
2012-06-11 8:24 pm
Difficulty: 50%Given that A₁ = (0, a₁), A₂ = (0, a₂) and A₃ = (0, a₃) are points on the y -axis, B₁, B₂and B₃ are points on thecurve x² + y² = 1 and P = (- 1, 0), suchthat P, A₁ and B₁arecollinear, P, A₂ and B₂ are collinear, P, A₃ and B₃ are collinear and B₁B₂ areparallelto PB₃. Expressa₃ in terms of a₁ and a₂.
回答 (1)
2012-06-12 1:22 am
✔ 最佳答案
In general, if P(-1, 0), A₁(0, a₁), and B₁ are collinear, then the coordinates of B₁ are ((1 - a₁²)/(1 + a₁²), 2a₁/(1 + a₁²))Therefore, the coordinates of B₂ and B₃ are ((1 - a₂²)/(1 + a₂²), 2a₂/(1 + a₂²)) and ((1 - a₃²)/(1 + a₃²), 2a₃/(1 + a₃²)) respectively.
As B₁B₂ // PB₃, ie slope of B₁B₂ = slope of PB₃, therefore,
(2a₁/(1 + a₁²) – 2a₂/(1 + a₂²))/[(1 - a₁²)/(1 + a₁²) - (1 - a₃²)/(1 + a₃²)] = [2a₃/(1 + a₃²) – 0]/[(1 - a₃²)/(1 + a₃²) – (-1)]
==> 2(a₁ - a₂)(1 - a₁a₂)/[(a₁ - a₂)(a₁ + a₂) = 2a₃/2
==> a₃ = (a₁a₂ - 1)/(a₁ + a₂)
2012-06-11 17:26:10 補充：
post 咗之後先發覺打得好醜陋, sorry.
2012-06-11 20:31:57 補充：
唔好意思, 打錯咗, 成段打過.
[2a₁/(1 + a₁²) – 2a₂/(1 + a₂²)]/[(1 - a₁²)/(1 + a₁²) - (1 - a₂²)/(1 + a₂²)] = [2a₃/(1 + a₃²) – 0]/[(1 - a₃²)/(1 + a₃²) – (-1)]
==> 2(a₁ + a₁a₂² - a₂ - a₁²a₂)/(-2a₁² + 2a₂²) = 2a₃/2
==> (a₁ - a₂)(1 - a₁a₂)/[( a₂ - a₁)( a₂ + a₁)] = 2a₃/2
==> a₃ = (a₁a₂ - 1)/(a₁ + a₂)
2012-06-11 20:32:21 補充：
手寫之後打落機景然2,3倒轉打.1,2又倒轉打. 以為答案冇錯就算, Sorry.
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