急 ! F5 數 Locus 9q1

(7) Let the pool side line be the x - axis.
Let the status of a fish be on the y - axis and at a distance of k from the pool side. So co - ordinates of the status is ( 0, - k).
Let co - ordinates of the boat be (x, y).
So distance of the boat to the pool side = - y.
Distance between the boat and the status
= sqrt [( x - 0)^2 + (y - k)^2]
Since the distance is always the same,
so sqrt [ (x - 0)^2 + (y - k)^2] = - y
x^2 + (y - k)^2 = y^2
x^2 + y^2 - 2ky + k^2 = y^2
2ky = x^2 + k^2
y = (1/2k)(x^2 + k^2) which is a parabola with axis of symmetry x = 0, that is the y - axis = the perpendicular line passing through the status and the pool side.
(8)
Let ABC be the y - axis and line CL be the x - axis.
Let B be ( 0, b), so A is (0, 2b).
Let P be (x, y)
Distance of AP = sqrt [(x - 0)^2 + (y - 2b)^2]
Distance from P to CL = y
Since the 2 distances are equal, so
sqrt[(x - 0)^2 + (y - 2b)^2] = y^2
x^2 + y^2 - 4by + 4b^2 = y^2
4by = x^2 + 4b^2
y = x^2/4b + b
which is a parabola with vertex (0, b) that is point B.
Since b is positive, so the parabola is opening upward.




2012-12-14 08:44:09 補充：
Correction : For Q7, the distance between the boat and the status should be
sqrt [ (x - 0)^2 + ( y + k)^2] because the co - ordinates of the fish is ( 0, - k). So the locus should be y = - x^2/2k - k/2. ( opening downward).

2012-12-14 08:45:47 補充：
Vertex is (0, - k/2)