math locus

1.
(a)
P(x, y) is equidistant from (-3, 2) and L : x = -5 :
√[(x + 3)² + (y - 2)²] = |x + 5|
(x + 3)² + (y - 2)² = |x + 5|²
x² + 6x + 9 + y² - 4y + 4 = x² + 10x + 25
C : y² - 4x - 4y - 12 = 0

(b)(i)
y axis : x = 0

Put x = 0 into C :
y² - 4y - 12 = 0

Discriminant of the above equation, Δ
= (-4)² - 4(1)(-12)
= 64 > 0

There are two distinct roots of y in the above equation and thus C meets they-axis at two distinct points.

(b)(ii)
y² - 4y - 12 = 0
(y + 6)(y - 2) = 0
y = -6 or y = 2

A = (0, -6), B = (0, 2)

(c)
Let (x, y) be the coordinates of N.

ΔBMN and ΔABM have the same height from B.
Area of ΔBMN : Area of ABM = MN : AM
Hence, MN : AM = 2 : 5
(x + 5) : (0 + 5) = 2 : 5
5x + 25 = 10
5x = -15
x = -3

Put x = -3 into C :
y² - 4(-3) - 4y - 12 = 0
y² - 4y = 0
y(y - 4) = 0
y = 0 or y = 4 (rejected)

The coordinates of N = (-3, 0)


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2.
(a)(i)
Let (a, b) be the coordinates of A.

√[(a - 0)² + (b - 0)²] = √[(a - 2)² + (b - 4)²] ... [1]
√[(a - 0)² + (b - 0)²] = √[(a + 6)² + (b - 0)²] ... [2]

From [1] :
a² + b² = a² - 4a + 4 + b² - 8b + 16
a + 2b - 5 = 0 ... [3]

From [2] :
a² + b² = a² + 12a + 36 + b²
a = -3

Put a = -3 into [3] :
(-3) + 2b - 5 = 0
b = 4

Equation of the locus of P :
√[(x + 3)² + (y - 4)²] = √[(0 + 3)² + (0 - 4)²]
x² + 6x + 9 + y² - 8y + 16 = 9 + 16
x² + y² + 6x - 8y = 0

(a)(ii)
From (a)(i), the coordinates of A = (-3, 4)

(b)(i)
The locus of Q is two concentric circle of the locus of P. One is outside oflocus of P and another is inside of locus of P.

(b)(ii)
AB = √[((3/5) + 3)² + ((44/5) - 4)²] = 6
AC = √[((1/5) + 3)² + ((8/5) - 4)²] 4

PB = PC = 1
Hence, both lie on the locus of Q at the same time.

2015-01-31 11:00:38 補充：
3.
在意見欄

2015-01-31 11:01:29 補充：
3.
(a)
Let (a, a) be the centre of the circle.

(x - a)² + (y - a)² = a²
x² - 2ax + a² + y² - 2ay + a² = a²
x² + y² - 2ax - 2ay + a² = 0 ... [1]

2015-01-31 11:01:36 補充：
Put x = 1 and y = 2 into [1] :
(1)² + (2)² - 2a(1) - 2a(2) + a² = 0
a² - 6a + 5 = 0
(a - 1)(a - 5) = 0
a = 1 or a = 5

Equation of A :
x² + y² - 2(1)x - 2(1)y + (1)² = 0
x² + y² - 2x - 2y + 1 = 0

Equation of B :
x² + y² - 2(5)x - 2(5)y + (5)² = 0
x² + y² - 10x - 10y + 25 = 0

2015-01-31 11:01:52 補充：
(b)
L : y - 1 = m(x -1)
L : mx - y + (1 - m) = 0

If B and L do not interest :
Distance between L and the centre of B > 5
|m(5) - (5) + (1 - m)| > 5
|4m - 4| > 5
4m - 4 > 5 or -(4m - 4) > 5
4m > 9 or -4m > 1
m > 9/4 or m < -1/4

2015-02-02 23:23:08 補充：
其 實 是 不 可 以 的 。

其實可唔可以話
The locus of Q is a sector with centre A, radius 10, subtended angle 120 degrees.