For any real value of A, P is the point (-h-2rCosA, k+2rSinA), where h,k and r are constants.
As A varies, find the equation of the locus of P.
locus problems
2008-06-17 3:09 am
回答 (2)
2008-06-17 3:14 am
✔ 最佳答案
P (-h - 2rcosA , k + 2rsinA)That is:
x = -h - 2rcosA
x + h = -2rcosA ─── (1)
y = k + 2rsinA
y - k = 2rsinA ─── (2)
(1)2 + (2)2:
(x + h)2 + (y - k)2 = 4r2(cos2A + sin2A)
Equation of locus:
(x + h)2 + (y - k)2 = 4r2
2008-06-16 20:44:54 補充：
There is another expression for the equation of locus:
(x + h)^2 + (y - k)^2 = 4r^2
x^2 + y^2 + 2xh - 2yk + (h^2 + k^2 - 4r^2) = 0
The locus is a circle.
參考: Myself~~~
2008-06-17 3:21 am
咁得唔得...
let x=-h-2rCosA, y=k+2rSinA
x=-h-2rCosA
cosA=x+h/(-2r)___(1)
y=k+2rSinA
sinA=y-k/2r___(2)
(cosA)^2 (sinA)^2=1 ---用恆等式
(1)^2+(2)^2=1
[x h/(-2r)]^2+[y-k/(2r)]^2=1
慢慢砌......
let x=-h-2rCosA, y=k+2rSinA
x=-h-2rCosA
cosA=x+h/(-2r)___(1)
y=k+2rSinA
sinA=y-k/2r___(2)
(cosA)^2 (sinA)^2=1 ---用恆等式
(1)^2+(2)^2=1
[x h/(-2r)]^2+[y-k/(2r)]^2=1
慢慢砌......
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