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Difficulty: MediumLoci and Geometric Constructions

Find the equation of the locus of a point P(x,y)P(x, y) that moves such that its distance from the fixed point (3,0)(3, 0) is always equal to its perpendicular distance from the vertical line x=−3x = -3.

Answer: y^2 = 12x / y^2 - 12x = 0 / y^2=12x / y^2 - 12x = 0

Answer

The equation of the locus is y2=12xy^2 = 12x (or y2−12x=0y^2 - 12x = 0).
Equating the distance from P(x,y)P(x, y) to (3,0)(3, 0), which is (x−3)2+y2\sqrt{(x - 3)^2 + y^2}, and the distance from P(x,y)P(x, y) to x=−3x = -3, which is ∣x+3∣|x + 3|, squaring both sides gives x2−6x+9+y2=x2+6x+9x^2 - 6x + 9 + y^2 = x^2 + 6x + 9. Subtracting x2+9x^2 + 9 from both sides yields y2=12xy^2 = 12x.

Step-by-Step Solution

1
Express the distance from P(x,y)P(x, y) to the point (3,0)(3, 0) using the distance formula.
d1=(x−3)2+(y−0)2=(x−3)2+y2d_1 = \sqrt{(x - 3)^2 + (y - 0)^2} = \sqrt{(x - 3)^2 + y^2}
The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
2
Express the perpendicular distance from P(x,y)P(x, y) to the line x=−3x = -3.
d2=∣x−(−3)∣=∣x+3∣d_2 = |x - (-3)| = |x + 3|
The perpendicular distance from a point (x,y)(x, y) to a vertical line x=kx = k is given by ∣x−k∣|x - k|.
3
Set the two distance expressions equal according to the locus condition.
(x−3)2+y2=∣x+3∣\sqrt{(x - 3)^2 + y^2} = |x + 3|
The locus condition states that the distance to (3,0)(3, 0) is equal to the distance to line x=−3x = -3.
4
Square both sides and simplify to obtain the Cartesian equation.
(x−3)2+y2=(x+3)2  ⟹  x2−6x+9+y2=x2+6x+9  ⟹  y2=12x(x - 3)^2 + y^2 = (x + 3)^2 \implies x^2 - 6x + 9 + y^2 = x^2 + 6x + 9 \implies y^2 = 12x
Squaring eliminates the square root and absolute value signs, leading to the algebraic representation of the locus.

Key Concept

Definition and equation of a parabola as the locus of a point equidistant from a fixed point (focus) and a fixed line (directrix)

Alternative Method

Recognize that the definition of a parabola is the locus of points equidistant from a focus (a,0)(a, 0) and a directrix x=−ax = -a. Here a=3a = 3, so the standard equation y2=4axy^2 = 4ax directly gives y2=4(3)x=12xy^2 = 4(3)x = 12x.
Estimated Time:1m 30s
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Topics

Loci and Geometric Constructions

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