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Analytic Geometry – Problem 35: Find the locus of points in the plane that are equidistant from the point and the…

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Question

Find the locus of points (x,y)(x,y) in the plane that are equidistant from the point F(0,−3)F(0,-3) and the line y=3y=3.

Step-by-step solution

(1) A point P(x,y)P(x,y) is on the locus iff its distance to the focus equals its distance to the directrix: PF=dist(P, y=3).PF=\text{dist}(P,\,y=3). So x2+(y+3)2=∣y−3∣.\sqrt{x^2+(y+3)^2}=|y-3|. (2) Squaring both sides: x2+(y+3)2=(y−3)2.x^2+(y+3)^2=(y-3)^2. Expand and simplify: x2+y2+6y+9=y2−6y+9⇒x2+12y=0.x^2+y^2+6y+9=y^2-6y+9\Rightarrow x^2+12y=0. Thus the locus is the parabola x2=−12y.x^2=-12y.

Final answer

The locus is the parabola x2=−12yx^2=-12y.

Marking scheme

Step 1 — Setup

Checkpoint: set up the distance equality x2+(y+3)2=∣y−3∣\sqrt{x^2+(y+3)^2}=|y-3| (2 pts)

Step 2 — Key Calculation

Checkpoint: square correctly and simplify to x2+12y=0x^2+12y=0 (3 pts)

Step 3 — Final Answer

Checkpoint: present x2=−12yx^2=-12y as the locus equation (2 pts)

Zero credit if: forgets absolute value and produces an invalid locus.

Deductions: -1 pt for algebraic expansion error.

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