# Equation of the Locus of a Moving Point

**Definition of a Locus**

Locus is a Latin word which means "place". In mathematics, locus is the set of points that satisfies the same geometrical properties. In most cases, the relationship of these points is defined according to their position in rectangular coordinates.

**Example 1**

Determine the equation of the curve such that the sum of the distances of any point of the curve from two points whose coordinates are (-3, 0) and (3, 0) is always equal to 8.

A. 4x^{2} + 49y^{2} - 343 = 0

B. 7x^{2} + 16y^{2} - 112 = 0

C. 7x^{2} - 16y^{2} + 112 = 0

D. 4x^{2} - 49y^{2} + 343 = 0

**Solution**

$d_1 + d_2 = 8$

$d_1 = 8 - d_2$

${d_1}^2 = (8 - d_2)^2$

${d_1}^2 = 64 - 16d_2 + {d_2}^2$

$(x + 3)^2 + (y - 0)^2 = 64 - 16\sqrt{(x - 3)^2 + (y - 0)^2} + \left[ (x - 3)^2 + (y - 0)^2 \right ]$

$(x^2 + 6x + 9) = 64 - 16\sqrt{(x - 3)^2 + y^2} + (x^2 - 6x + 9)$

$12x - 64 = 16\sqrt{(x - 3)^2 + y^2}$

$3x - 16 = 4\sqrt{(x - 3)^2 + y^2}$

$(3x - 16)^2 = 16 \left[ (x - 3)^2 + y^2 \right]$

$9x^2 - 96x + 256 = 16 \left[ (x^2 - 6x + 9) + y^2 \right]$

$9x^2 - 96x + 256 = 16x^2 - 96x + 144 + 16y^2$

$7x^2 + 16y^2 - 112 = 0$ Answer [ B ]

**Another Solution**

The following solution is not under this topic. This can be done with your knowledge in conic sections, particularly ellipse.

## Click here to expand or collapse this section

From the figure above, c = 3

$2a = 8$

$a = 4$

$a^2 = 16$

$b^2 + c^2 = a^2$

$b^2 + 3^2 = 4^2$

$b^2 = 7$

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

$\dfrac{x^2}{16} + \dfrac{y^2}{7} = 1$

$7x^2 + 16y^2 = 16(7)$

$7x^2 + 16y^2 - 112 = 0$ Answer [ B ]

- Log in to post comments