6.6 Equation of a Locus

6.6 Equation of a Locus

1. The equation of the locus of a moving point P(x, y) which is always at a constant distance (r) from a fixed point (x₁, y₁) is:

2. The equation of the locus of a moving point P(x, y) which is always at a constant distance from two fixed points (x₁, y₁) and (x₁, y₁) with a ratio is:

3. The equation of the locus of a moving point P (x, y) which is always equidistant from two fixed points A and B is the perpendicular bisector of the straight line AB.

Example 1

Find the equation of the locus of a moving point P(x, y) which is always at a distance of 5 units from a fixed point Q (2, 4).

Solution:

(x – x₁)²+ (y – y₁)² = r²

(x – 2)² + (y – 4)² = 5²

x² – 4x + 4 + y² – 8y + 16 = 25

x² + y²– 4x – 8y – 5 = 0

Example 2

Find the equation of the locus of a moving point P(x, y) which is always equidistant from points A (-2, 3) and B (4, -1).

Solution:

\begin{array}{l} Given P A = P B \\ \sqrt{{(x - (- 2))}^{2} + {(y - 3)}^{2}} = \sqrt{{(x - 4)}^{2} + {(y - (- 1))}^{2}} \\ Square both sides to eliminate the square roots . \\ {(x + 2)}^{2} + {(y - 3)}^{2} = {(x - 4)}^{2} + {(y + 1)}^{2} \\ x^{2} + 2 x + 4 + y^{2} - 6 y + 9 = x^{2} - 8 x + 16 + y^{2} + 2 y + 1 \\ 10 x - 8 y - 4 = 0 \\ Hence, the equation of the locus of point P is \\ 10 x - 8 y - 4 = 0 \end{array}

Example 3

A (2, 0) and B (0, -2) are two fixed points. Point P moves with a ratio so that AP: PB = 1: 2. Find the equation of the locus of point P.

Solution:

\begin{array}{l} A P : P B = 1 : 2 \\ \frac{A P}{P B} = \frac{1}{2} \\ 2 A P = P B \\ 2 \sqrt{{(x - 2)}^{2} + {(y - 0)}^{2}} = \sqrt{{(x - 0)}^{2} + {(y - (- 2))}^{2}} \\ Square both sides to eliminate the square roots . \\ 4 [{(x - 2)}^{2} + y^{2}] = x^{2} + {(y + 2)}^{2} \\ 4 (x^{2} - 4 x + 4 + y^{2}) = x^{2} + y^{2} + 4 y + 4 \\ 4 x^{2} - 16 x + 16 + 4 y^{2} = x^{2} + y^{2} + 4 y + 4 \\ 3 x^{2} + 3 y^{2} - 16 x - 4 y + 12 = 0 \\ Hence, the equation of the locus of point P is \\ 3 x^{2} + 3 y^{2} - 16 x - 4 y + 12 = 0 \end{array}