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JEE Mains Maths · Complex Numbers

Arcs and Conic Loci

Loci set by the angle at z — a quotient that is real, purely imaginary or of fixed argument — and loci that are ellipses, hyperbolas or parabolas.

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Why this matters

Sixteen PYQs, thirteen of them multiple choice. Twelve fix the argument of a quotient of the form (z − a)/(z − b), which puts z on a line, a circle or an arc; four are conic sections, usually an ellipse from a sum of two distances. Two ideas cover the page.

Concept 1 of 2: Real, imaginary or fixed-argument quotients

arg⁡z−az−b\arg\frac{z-a}{z-b} is the angle at zz between the directions to bb and to aa. If the quotient is real, that angle is 0 or π\pi: zz is on the line through aa and bb. If it is purely imaginary, the angle is a right angle, so zz is on the circle with diameter abab. If the argument is a fixed θ\theta, zz runs along one arc of a circle through aa and bb on which the chord abab subtends θ\theta.

Definition

  • Real: the line through aa and bb, without bb.
  • Purely imaginary: the circle with diameter abab, without aa and bb.
  • Argument θ\theta: one arc through aa and bb, radius ∣a−b∣2sin⁡θ\frac{|a-b|}{2\sin\theta}, centre on the perpendicular bisector.
  • Find which arc by testing a point, such as one on the bisector.

Radius of the arc

arg⁡z−az−b=θ ⇒ r=∣a−b∣2sin⁡θ\arg\frac{z-a}{z-b}=\theta\ \Rightarrow\ r=\frac{|a-b|}{2\sin\theta}

Worked example

Describe the locus arg⁡z−3z+3=π2\arg\frac{z-3}{z+3}=\frac\pi2.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q154Moderate

Example 1 · Complex Numbers · Arcs and Conic Loci

Let zz be a complex number such that the real part of z−2iz+2i\frac{z- 2i}{z+ 2i} is zero. Then, the maximum value of ∣z−(6+8i)∣|z - (6 + 8i)| is equal to:

A fixed argument gives one arc

arg⁡z−az−b=θ\arg\frac{z-a}{z-b}=\theta is one arc, not the whole circle; the other arc has the argument θ−π\theta-\pi. Before counting intersections or measuring distances, check that the point found lies on the right arc.
Drill 11 more on real, imaginary or fixed-argument quotients

Concept 2 of 2: Ellipses, hyperbolas and parabolas

∣z−a∣+∣z−b∣=2k|z-a|+|z-b|=2k is an ellipse with foci aa and bb when 2k>∣a−b∣2k>|a-b|; it is only the segment abab when 2k=∣a−b∣2k=|a-b|, and empty when 2k2k is smaller. A difference ∣∣z−a∣−∣z−b∣∣=2k<∣a−b∣\big||z-a|-|z-b|\big|=2k<|a-b| is a hyperbola. ∣z−a∣=Re(z)|z-a|=\mathrm{Re}(z) sets a distance to a point equal to a distance to a line: a parabola. The focus facts often answer a question without any equation.

Definition

  • ∣z−a∣+∣z−b∣=2k>∣a−b∣|z-a|+|z-b|=2k>|a-b|: ellipse, with c=∣a−b∣2c=\frac{|a-b|}2 and semi-minor axis k2−c2\sqrt{k^2-c^2}.
  • 2k=∣a−b∣2k=|a-b|: the segment abab.
  • ∣∣z−a∣−∣z−b∣∣=2k<∣a−b∣\big||z-a|-|z-b|\big|=2k<|a-b|: hyperbola.
  • Distance to a point equal to distance to a line: parabola.
  • The least value of ∣z−a∣+∣z−b∣|z-a|+|z-b| is ∣a−b∣|a-b|.

Ellipse from two foci

∣z−z1∣+∣z−z2∣=2a>∣z1−z2∣: b2=a2−c2, c=12∣z1−z2∣|z-z_1|+|z-z_2|=2a>|z_1-z_2|:\ b^2=a^2-c^2,\ c=\tfrac12|z_1-z_2|

Worked example

Describe ∣z−3∣+∣z+3∣=10|z-3|+|z+3|=10.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q52Moderate

Example 2 · Complex Numbers · Arcs and Conic Loci

The number of values of z∈Cz\in\mathbb{C}, satisfying the equations ∣z−(4+8i)∣=10|z- (4 + 8i)| =\sqrt{10} and ∣z−(3+5i)∣+∣z−(5+11i)∣=45|z- (3 + 5i)| + |z- (5 + 11i)| = 4\sqrt{5}, is :

Compare the constant with the focal distance

A sum of distances is an ellipse only when the constant exceeds the distance between the two points. Check 2k2k against ∣a−b∣|a-b| before using aa, bb and cc.
Drill 3 more on ellipses, hyperbolas and parabolas

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Real, imaginary or fixed-argument quotients

    Radius of the arc

    arg⁡z−az−b=θ ⇒ r=∣a−b∣2sin⁡θ\arg\frac{z-a}{z-b}=\theta\ \Rightarrow\ r=\frac{|a-b|}{2\sin\theta}
  • Ellipses, hyperbolas and parabolas

    Ellipse from two foci

    ∣z−z1∣+∣z−z2∣=2a>∣z1−z2∣: b2=a2−c2, c=12∣z1−z2∣|z-z_1|+|z-z_2|=2a>|z_1-z_2|:\ b^2=a^2-c^2,\ c=\tfrac12|z_1-z_2|

Watch out for (2)

Test yourself on Complex Numbers

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Complex Numbers chapter testDrill all 16 questions