A point P(x, y) moves in a Cartesian plane such that the square of its distance from A(3, … — JAMB UTME

A point $P(x, y)$ moves in a Cartesian plane such that the square of its distance from $A(3, 0)$ exceeds the square of its distance from $B(-1, 2)$ by $4$ units. Which of the following equations represents the locus of $P$ ?

$2x - y = 0$ Answer
B
$2x + y - 4 = 0$
C
$x - 2y + 1 = 0$
D
$x^2 + y^2 - 4x + 2y - 4 = 0$

Answer

The equation representing the locus of

P

2x - y = 0

Using the distance formula,

PA^2 = (x - 3)^2 + y^2 = x^2 - 6x + 9 + y^2

and

PB^2 = (x + 1)^2 + (y - 2)^2 = x^2 + 2x + 1 + y^2 - 4y + 4

. Subtracting

PB^2

from

PA^2

yields

(x^2 + y^2 - 6x + 9) - (x^2 + y^2 + 2x - 4y + 5) = 4

, which simplifies to

-8x + 4y = 0

, or

2x - y = 0

Step-by-Step Solution

Express the square of the distance from

P(x, y)

A(3, 0)

using the distance formula

PA^2 = (x - 3)^2 + (y - 0)^2 = x^2 - 6x + 9 + y^2

The distance formula gives

PA^2 = (x - x_A)^2 + (y - y_A)^2

Express the square of the distance from

P(x, y)

B(-1, 2)

PB^2 = (x - (-1))^2 + (y - 2)^2 = (x + 1)^2 + (y - 2)^2 = x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 + 2x - 4y + 5

Expand both algebraic squares for

PB^2

Set up the locus condition

PA^2 - PB^2 = 4

and simplify

(x^2 + y^2 - 6x + 9) - (x^2 + y^2 + 2x - 4y + 5) = 4 \implies -8x + 4y + 4 = 4

Subtracting

PB^2

from

PA^2

cancels the quadratic

x^2

and

y^2

terms.

Rearrange into standard linear equation form

-8x + 4y = 0 \implies 2x - y = 0

Divide the entire equation by

-4

to simplify.

Key Concept

Locus defined by difference of squares of distances to two fixed points

Estimated Time:1m 30s

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