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Visual SolutionPYQ 2024 · Jan Shift 1Tricky

Question

z

satisfies

|z - 1| + |z + 1| = 4

, then the eccentricity of the locus of

z

is:

(A)

\frac{1}{2}

(B)

\frac{\sqrt{3}}{2}

(C)

\frac{1}{\sqrt{2}}

(D)

\frac{1}{3}

Solution Path

Recognize

|z-1| + |z+1| = 4

as ellipse definition with foci

(\pm1, 0)

, so

c = 1

a = 2

e = c/a = 1/2

01Question Setup

1/4

We need to find the eccentricity of the locus defined by

|z - 1| + |z + 1| = 4

. This is a sum of distances from two fixed points.

|z - 1| + |z + 1| = 4

means the sum of distances from

(1,0)

and

(-1,0)

is constant

02Identify the ConicKEY INSIGHT

2/4

A point whose sum of distances from two fixed points (foci) is constant traces an ellipse. Here foci are at

F_1(1,0)

and

F_2(-1,0)

, and the constant sum is 4.

2a = 4 \Rightarrow a = 2

, foci at

(\pm 1, 0)

c = 1

03Compute Eccentricity

3/4

For an ellipse, eccentricity

e = c/a

where

c

is the distance from center to focus and

a

is the semi-major axis.

e = \frac{c}{a} = \frac{1}{2}

04Final Answer

4/4

The ellipse has semi-major axis

a = 2

, semi-minor axis

b = \sqrt{a^2 - c^2} = \sqrt{3}

, and eccentricity

1/2

Eccentricity

= \frac{1}{2}

, Answer: (A)

Concepts from this question1 concepts unlocked

★

Conjugate Symmetry on Argand Plane

STANDARD

Conjugate pairs are reflections across the real axis

z = x + iy \implies \bar{z} = x - iy

Recognizing conjugate pairs halves the computation - products simplify dramatically

Locus problemsGeometric interpretationSymmetric root problems

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