What is the equation of the locus of a point P(x, y) that is always equidistant from the p… — JAMB UTME

What is the equation of the locus of a point $P(x, y)$ that is always equidistant from the point $(0, 4)$ and the line $y = -4$ ?

Answer: x^2 = 16y / x^2 - 16y = 0 / x^2-16y=0 / y = x^2/16 / y = \frac{x^2}{16} / x^{2}=16y / x^{2}-16y=0

The equation of the locus is

x^2 = 16y

(or

x^2 - 16y = 0

Equating the distance from

P(x, y)

(0, 4)

, which is

\sqrt{x^2 + (y-4)^2}

, to the perpendicular distance to the line

y = -4

, which is

|y + 4|

, and squaring both sides gives

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

. Expanding yields

x^2 + y^2 - 8y + 16 = y^2 + 8y + 16

, which simplifies directly to

x^2 = 16y

x^2 - 16y = 0

Formulate the distance expressions from point

P(x, y)

to the given point

(0, 4)

and line

y = -4

Distance to

(0, 4) = \sqrt{(x - 0)^2 + (y - 4)^2}

. Perpendicular distance to

y = -4

|y - (-4)| = |y + 4|

By definition of geometric locus, the distance from

P(x, y)

to the fixed point must equal its distance to the fixed line.

Equate the two distance expressions.

\sqrt{x^2 + (y - 4)^2} = |y + 4|

The point

P(x, y)

is equidistant from both geometric entities.

Square both sides of the equation to clear the square root and absolute value.

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

Squaring eliminates radical and absolute value signs while preserving algebraic equality.

Expand both squared binomial expressions and simplify.

x^2 + y^2 - 8y + 16 = y^2 + 8y + 16 \implies x^2 = 16y

Subtracting

y^2 + 16

from both sides leaves

x^2 - 8y = 8y

, which simplifies to

x^2 = 16y

The locus of points equidistant from a fixed point (focus) and a fixed straight line (directrix) forms a parabola.

Estimated Time:1m 30s

Question