MT_F4_7.4 Equation of Locus

7.4 Equation of Locus (Loci)

路线

- The locus of a point P (x, y) is the track (or path) of the point.

- It is moving subject to a certain condition.

- The locus of a moving point can be represented by an equation.

- Two types of conditions: 定数

• Locus of a moving point from a fixed point is a constant.

• Ratio of distance of moving point from two fixed points is a constant

比例

Example：

1. Find the equation of the locus of a moving point P such that its distance

from the point A (2, 4) is 2 units. AP = 2 units

A2,4 Px, y Distance
(x2  x1)2  ( y2  y1)2
AP  2
(x2  x1)2  ( y2  y1)2  2 Square both sides
of the equation
(x  2)2  ( y  4)2  2
(x  2)2  ( y  4)2  22
x2  4x  4  y2  8y 16  4

x2  y2  4x 8y 16  0

The equation of locus of P is x2  y2  4x 8y 16  0

2. Find the equation of the locus of a moving point Q such that its distances
from the points K (3, –7) and H (–5, 1) are equal.

Qx, y , K 3,7 Qx, y , H  5,1 PK = PH

QK = QH

(x  3)2  ( y  7)2  (x  5)2  ( y 1)2 Square both sides
of the equation
(x  3)2  ( y  7)2  (x  5)2  ( y 1)2

x2  6x  9  y2 14y  49  x2 10x  25  y2  2y 1

16x 16y  32  0

xy2  0

The equation of locus of Q is x  y  2  0

3. P (2, 0) and Q (0, –2) are two fixed points. The point T moves in such a
way that such PT : TQ  2 : 3 . Find the equation of the locus of point T.

T x, y , P2,0 PT :TQ  2 : 3 Ratio

PT  2 T x, y , Q0,2
TQ 3
3 PT  2 TQ

3 (x  2)2  ( y  0)2  2 (x  0)2  ( y  2)2 Square both sides

   32 (x  2)2  y2  22 x2  ( y  2)2 of the equation

9(x2  4x  4  y2 )  4(x2  y2  4 y  4)

9x2  36 x  36  9 y2  4x2  4 y2 16 y 16

5x2  5y2  36x 16y  20  0

The equation of the locus of point T is 5x2  5y2  36x 16y  20  0

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3. Point Q moves along the arc of a circle with centre (6, 5). The arc of the
circle passes through R (2, 8) and S (k, 2). Find
(a) the equation of the locus of Q

(b) the values of k - Draft the situation
- a circle
R (2, 8) - R and S are on the arc
- locus Q moves along
·O the arc

(6, 5)

S (k, 2)

Q moves along the arc

(a) OR is the radius of circle Distance
O (6, 5) R (2, 8) (x2  x1)2  ( y2  y1)2

OR  (6  2)2  (5  8)2
 25
 5 units

OQ  5 OQ = OR

(x  6)2  ( y  5)2  5

(x  6)2  ( y  5)2  52
x2 12x  36  y2 10y  25  25

x2  y2 12x 10y  36  0

The equation of locus of Q is x2  y2 12x 10y  36  0

(b) S (k, 2)
x2  y2 12x 10y  36  0

k 2  22 12k 102 36  0

k 2  4 12k  20  36  0
k2 12k  20  0

k  2k 10  0

k  2 or k  10

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