concept of locus

1. Let P be (x,y)
方法係 : distance from x-axis to P = distance from Q to P
右邊係 distance between two points, 用公式，應該識啦?
左邊呢? distance from point to line, 即係取垂直線的長度，
你諗下先，再睇答案
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distance from x-axis = y units
distance from Q = sqrt( ( y-4)^2 + x^2 )
locus is : y = sqrt( ( y-4)^2 + x^2 )
y^2 = y^2 - 8 y + 16 + x^2
y = 2 + x^2 / 8

係一條parabola，Q 叫 focus, x-axis 是 focal plane

2.
睇圖先: http://www.sendspace.com/file/uu8yof
4 fix, r fix, 即係 R 都fix la~
其實答案一個圓，搵 r 先, Pyth Theorem, 再搵埋R, 再 equantion of circle
see? Try it!
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x^2+y^2-4x+6y+1=0
(x-2)^2 + (y+3)^2 -4 -9 +1 = 0
center = (2,-3) ; r = sqrt(12)
R = sqrt( 12+4^2) = sqrt(28)
required equation is (x-2)^2 + (y+3)^2 -28 = 0

Q1.
square of distance from P(x,y) to Q(0,4) = (x-0)^2 + (y-4)^2 = x^2 + (y-4)^2.
square of distance from P(x,y) to x-axis = y^2. Therefore,
x^2 + (y-4)^2 = y^2
x^2 + y^2 + 16 - 8y = y^2
8y = x^2 + 16 is the locus of P.
Q.2
Centre of circle is (2,-3), radius = sqrt(12).
square of distance from centre to P(x,y) = (x-2)^2 + (y+3)^2.
square of length of tangent = 4^2 = 16.
square of radius = 12. Therefore, by Pythagoras theorem,
(x-2)^2 + (y+3)^2 = 16 + 12
x^2 + 4 -4x + y^2 + 9 +6y -28 = 0.
Therefore, locus of P is x^2 + y^2 -4x +6y -15 = 0.