1. P(x,y) is a variable point such that the distance from P to the line x-4=0 is always equal to twice the distance between P and the point (1,0).
Show that the equation of the locus of P is 3x^2+4y^2-12=0
2. Solve the following inequalities:
a. lx-1l>2
b. llyl-1l>2
數學2002pp
2011-09-25 1:24 am
回答 (1)
2011-09-25 1:41 am
✔ 最佳答案
1 The distance from P to L: x - 4= 0 is (x - 4)The distancefrom P to (1,0) is √[(x - 1)^2 + y^2]
So, x - 4 = 2√[(x - 1)^2 + y^2]
(x - 4)^2 = 4[(x - 1)^2 + y^2]
x^2 - 8x + 16 = 4x^2 - 8x + 4 + 4y^2
3x^2 + 4y^2 - 12 = 0
2(a) |x - 1| > 2
x - 1 < - 2 or x - 1 > 2
x < - 1 or x > 3
(b) ||y| - 1| > 2
|y| < -1 (rejected) or |y| > 3 by the result of (a)
So, y < - 3 or y > 3
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