Locus
A moving point keeps its distance from the point (4, 0) always equal to its distance from the vertical line x = -4; what curve does the point trace?
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About This Question
- Subject
- mathematics
- Chapter
- coordinate geometry
- Topic
- locus
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
y2=16x
The defining property of a parabola is that each of its points is equidistant from a fixed focus and a fixed directrix, so this verbal condition is exactly a parabola's locus definition. Let the moving point be (x, y); distance to the focus (4, 0) is sqrt((x - 4)^2 + y^2) and distance to the line x = -4 is |x + 4|. Equating and squaring: (x - 4)^2 + y^2 = (x + 4)^2. Expanding, x^2 - 8x + 16 + y^2 = x^2 + 8x + 16, so y^2 = 16x. Option y^2 = 8x corresponds to focus (2, 0) and directrix x = -2, not the given data. Option x^2 = 16y describes an upward parabola, inconsistent with a horizontal focus-directrix setup. Option y^2 = 4x uses a = 1 rather than 4. This is the foundational JEE Advanced locus-to-conic derivation. Plausibility check: comparing y^2 = 16x with y^2 = 4ax gives a = 4, matching the focus (4, 0) and directrix x = -4 exactly, so the derived curve is self-consistent.
This medium difficulty mathematics question is from the chapter coordinate geometry, covering the topic of locus. It appeared in the 2025 exam.
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