(A) The equation $|z-z_{1}|+|z-z_{2}|=k$ represents an ellipse if $k > |z_{1}-z_{2}|$.In the given equation,$k = 2|z_{1}-z_{2}|$.Since $2|z_{1}-z_{2}|

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The equation $|z-z_{1}|+|z-z_{2}|=k$ represents an ellipse if $k > |z_{1}-z_{2}|$.In the given equation,$k = 2|z_{1}-z_{2}|$.Since $2|z_{1}-z_{2}| > |z_{1}-z_{2}|$,the condition for an ellipse is satisfied.Therefore,the locus of $z$ is an ellipse with foci at $z_{1}$ and $z_{2}$.

Class 11 Mathematics 4-1.Complex numbers Geometry of complex numbers

Difficult

Let $z_{1}$ and $z_{2}$ be two fixed complex numbers in the Argand plane and $z$ be an arbitrary point satisfying $|z-z_{1}|+|z-z_{2}|=2|z_{1}-z_{2}|$. Then,the locus of $z$ will be

WBJEE 2014 All WBJEE

A
an ellipse
B
a straight line joining $z_{1}$ and $z_{2}$
C
a parabola
D
a bisector of the line segment joining $z_{1}$ and $z_{2}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

4-1.Complex numbers

Subtopic

Geometry of complex numbers

Previous Year

WBJEE 2014 Paper All WBJEE years →

If $z \neq 1$ and $\frac{z^2}{z-1}$ is real,then the point represented by the complex number $z$ lies:

DifficultAIEEE 2012

View Solution

If a complex number $z$ satisfies $|z|^2+1=|z^2-1|$,then the locus of $z$ is

EasyAP EAMCET 2018

View Solution

Let $A = \{z \in \mathbb{C} : |z - 2 - i| = 3\}$, $B = \{z \in \mathbb{C} : \operatorname{Re}(z - iz) = 2\}$ and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to . . . . . . .

DifficultJEE MAIN 2025

View Solution

Let $A(3-i)$ and $B(2+i)$ be two points in the Argand plane. If the point $P$ represents the complex number $z=x+iy$,which satisfies $|z-3+i|=|z-2-i|$,then the locus of the point $P$ is

DifficultAP EAMCET 2018

View Solution

If $a$ is a complex number and $b$ is a real number,then the equation $\bar{a}+a+b=0$ represents $a$ as a locus of points in the complex plane,which is a:

EasyAP EAMCET 2001

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Let $z_{1}$ and $z_{2}$ be two fixed complex numbers in the Argand plane and $z$ be an arbitrary point satisfying $|z-z_{1}|+|z-z_{2}|=2|z_{1}-z_{2}|$. Then,the locus of $z$ will be

Similar Questions

If $z \neq 1$ and $\frac{z^2}{z-1}$ is real,then the point represented by the complex number $z$ lies:

If a complex number $z$ satisfies $|z|^2+1=|z^2-1|$,then the locus of $z$ is

Let $A = \{z \in \mathbb{C} : |z - 2 - i| = 3\}$, $B = \{z \in \mathbb{C} : \operatorname{Re}(z - iz) = 2\}$ and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to . . . . . . .

Let $A(3-i)$ and $B(2+i)$ be two points in the Argand plane. If the point $P$ represents the complex number $z=x+iy$,which satisfies $|z-3+i|=|z-2-i|$,then the locus of the point $P$ is

If $a$ is a complex number and $b$ is a real number,then the equation $\bar{a}+a+b=0$ represents $a$ as a locus of points in the complex plane,which is a:

Vedclass Test Series

Exam Paper Generator

Online Exam Module