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Locus questions feel open-ended: there is no formula to plug into, and the answer could be a line, a circle or a parabola. In fact they follow one routine. This guide gives you a five-step method, works it on four common question types with every equation checked, and lists the slips that cost marks.
What is a locus?
A locus is the set of all points that satisfy a given condition. NCERT's Class 11 chapter Conic Sections defines each conic this way. A circle is the set of all points in a plane equidistant from a fixed point, and a parabola the set of all points equidistant from a fixed line and a fixed point. The same chapter's historical note records that Pierre de Fermat's treatise on plane and solid loci was published only after his death, in 1679.
For exams, the JEE (Main) 2026 syllabus lists "locus and its equation" in coordinate geometry, and the JEE (Advanced) 2026 brochure lists "locus problems". Check your year's documents.
The five-step method
Name the moving point (h, k). Use h and k, not x and y, so you do not mix it up with other points.
Write the condition as an equation in h, k and any given constants.
Eliminate any parameter (an angle, a slope, a length that varies).
Replace h by x and k by y.
Check for excluded points, where the condition does not make sense.
Example 1: equidistant from two points
Find the locus of a point equidistant from A(1, 2) and B(3, 4).
PA² = PB² gives (h − 1)² + (k − 2)² = (h − 3)² + (k − 4)².
Expand both sides: h² − 2h + 1 + k² − 4k + 4 = h² − 6h + 9 + k² − 8k + 16.
The squares cancel: −2h − 4k + 5 = −6h − 8k + 25, so 4h + 4k = 20, which gives h + k = 5.
Locus: x + y = 5.
Check: the midpoint of AB is (2, 3), and 2 + 3 = 5. The slope of AB is 1 and the slope of x + y = 5 is −1, so they are perpendicular. This is the perpendicular bisector, as geometry says it must be.
Example 2: a ratio of distances
A point P moves so that PA = 2PB, where A = (0, 0) and B = (3, 0). Find its locus.
PA² = 4PB²: h² + k² = 4[(h − 3)² + k²] = 4h² − 24h + 36 + 4k².
Collect terms: 3h² + 3k² − 24h + 36 = 0. Divide by 3: h² + k² − 8h + 12 = 0. Complete the square: (h − 4)² + k² = 4.
Locus: a circle with centre (4, 0) and radius 2.
Check two points on the x-axis. At (2, 0): PA = 2, PB = 1. At (6, 0): PA = 6, PB = 3. Both satisfy PA = 2PB.
Example 3: eliminating a parameter
A rod of length 10 slides with one end on the x-axis and the other on the y-axis. Find the locus of its midpoint.
Let the ends be (a, 0) and (0, b). The length condition is a² + b² = 100.
The midpoint is (h, k) = (a/2, b/2), so a = 2h and b = 2k.
Substitute: 4h² + 4k² = 100, so h² + k² = 25.
Locus: x² + y² = 25, a circle of radius 5 centred at the origin.
NCERT's Example 19 in the Conic Sections chapter uses the same set-up for a point that is not the midpoint and gets an ellipse. Try it after this one: the method is identical.
A parametric version. Find the locus of the midpoint of the segment joining the origin to a point (t², 2t) on the parabola y² = 4x. The midpoint is (h, k) = (t²/2, t). So t = k and h = k²/2, giving y² = 2x. Check with t = 2: the point is (4, 4), the midpoint (2, 2), and 2² = 2 × 2.
Example 4: watch for excluded points
A(−2, 0) and B(2, 0) are fixed. P moves so that angle APB is 90°. Find its locus.
Angle APB = 90° means PA is perpendicular to PB, so the slopes multiply to −1 (or use the dot product of vectors PA and PB):
(h + 2)(h − 2) + k² = 0, so h² + k² = 4.
Locus: the circle x² + y² = 4, excluding A and B. At A or B, the angle APB does not exist. This is the circle with diameter AB, as the angle-in-a-semicircle theorem predicts. Examiners give a mark for stating the exclusion.
What are the common mistakes?
Using x and y for everything. If a fixed point is (x1, y1) and the moving point is (x, y), the algebra tangles. Use (h, k).
Leaving the parameter in. An answer with t or θ in it is not a locus equation yet.
Forgetting excluded points, as in Example 4.
Squaring without care. Squaring both sides is safe for PA = PB, because distances are positive. With a condition such as "PA − PB = 2", squaring can add points where PB − PA = 2. Check a point.
Not simplifying to a standard form. Complete the square so you can name the curve.
A practice routine
Day 1: Six distance-condition loci (equidistant, ratio, sum of squares). Test a point on each.
Day 2: Four parametric loci: sliding rods, midpoints of chords, points on conics.
Day 3: Three loci with excluded points or angle conditions.
Day 5: A mixed timed set from previous JEE papers.
Many locus answers are circles or conics, so revise circles in coordinate geometry and our method for conic sections alongside this. If you cannot see what shape a condition traces, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Try it now: find the locus of a point whose distance from (0, 0) is three times its distance from (4, 0), then test one point.
