標題:
a.maths
發問:
最佳解答:
As follows~~~ 圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/parameter2.jpg?t=1189685023
其他解答:
(a) x=4at --------(1) , y=2at^2-1 -----------(2) from (1) x/4a = t sub into Eq(2): y = 4a(x/2a)^2 - 1 a(y+1) = x^2 (b)x= -a cos 2q ------(1), y=b sin 2q-----------(2) cos 2q = (-x/a) ---------(3) sin 2q = (y/b) ---------(4) Eq(3)^2 +Eq(4)^2: (x/a)^2+(y/b)^2 = (cos2q)^2 + (sin 2q)^2 = 1 (x^2/a^2) + (y^2/b^2) = 1 (c) x=a csc q ----------(1),y=b cot q ---------(2) x /a = csc q ---------(3) y/b = cot q ----------(4) Eq (3)^2 - Eq(4)^2: (x/a)^2 - (y/b)^2 = (csc q)^2 +(cot q)^2 = 1 (x^2 / a^2) - (y^2 / b^2) = 1
a.maths
發問:
此文章來自奇摩知識+如有不便請留言告知
From each of the following parametric equations, find an equation containing x and y by eliminating t or q. (a) x=4at, y=2at^2-1 (b)x= -a cos 2q, y=b sin 2q (c)x=a csc q ,y=b cot q最佳解答:
As follows~~~ 圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/parameter2.jpg?t=1189685023
其他解答:
(a) x=4at --------(1) , y=2at^2-1 -----------(2) from (1) x/4a = t sub into Eq(2): y = 4a(x/2a)^2 - 1 a(y+1) = x^2 (b)x= -a cos 2q ------(1), y=b sin 2q-----------(2) cos 2q = (-x/a) ---------(3) sin 2q = (y/b) ---------(4) Eq(3)^2 +Eq(4)^2: (x/a)^2+(y/b)^2 = (cos2q)^2 + (sin 2q)^2 = 1 (x^2/a^2) + (y^2/b^2) = 1 (c) x=a csc q ----------(1),y=b cot q ---------(2) x /a = csc q ---------(3) y/b = cot q ----------(4) Eq (3)^2 - Eq(4)^2: (x/a)^2 - (y/b)^2 = (csc q)^2 +(cot q)^2 = 1 (x^2 / a^2) - (y^2 / b^2) = 1
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