
The equation $\left.\left.\sqrt{(}(x-2)^{2}+y^{2}\right)+\sqrt{(}(x+2)^{2}+y^{2}\right)=4$ represents a/an
1) circle
2) straight lines
3) parabola
4) ellipse
Answer
164.1k+ views
Hint: In this question, first we square both sides of the given equation . After that we have to simplify the resultant expression to find out the y value. And according to the resultant y value, we can find out what the given expression represents.
Complete step by step Solution:
Given equation is
$\left.\left.\sqrt{(}(x-2)^{2}+y^{2}\right)+\sqrt{(}(x+2)^{2}+y^{2}\right)=4$
$\left.\left.\sqrt{(}(x-2)^{2}+y^{2}\right)=4-\sqrt{(}(x+2)^{2}+y^{2}\right)$
Squaring both sides
$\left.(x-2)^{2}+y^{2}=16+\left((x+2)^{2}+y^{2}\right)-8 \sqrt{(}(x+2)^{2}+y^{2}\right)$
$\left.\left(x^{2}+4-4 x+y^{2}\right)=16+x^{2}+4+4 x+y^{2}-8 \sqrt{(}(x+2)^{2}+y^{2}\right)$
Simplify the terms
$\left.-8 x=16-8 \sqrt{(}(x+2)^{2}+y^{2}\right)$
$\left.x=-2+\sqrt{(}(x+2)^{2}+y^{2}\right)$
$\left.(x+2)=\sqrt{(}(x+2)^{2}+y^{2}\right)$
Squaring again
$(x+2)^{2}=(x+2)^{2}+y^{2}$
$=>y^{2}=0$
$=>y=0$
Here the $y$ axis value is 0
That means there is no line that passes through the $y$ axis
This implies that the line passes through $x$ axis and it will be a straight line.
$=>x$ axis
The location of all the points in a plane whose sum of the distances from two fixed points in the plane is constant is known as an ellipse. The foci (singular focus), which are fixed points and are encircled by the curve, are known. Directrix is the fixed line, and the eccentricity of the ellipse is the constant ratio.
Then the ellipse, circle, or parabola will not be formed by this equation
Hence, the correct option is 2.
Additional information:
We know for a straight line value comes out to be zero.
And A circle is a collection of all points in a plane that is uniformly distanced from a fixed point. The radius of a circle is the separation between the center and any point along its circumference.
we can arrive at the following parabola equations:
$y^{2}=4 a x$
$y^{2}=-4 a x$
$x^{2}=4 a y$
$x^{2}=-4 a y$
The term "standard equations of parabolas" refers to these four equations. It is significant to note that one of the coordinate axes, the vertex at the origin, is the focus of the common equations for parabolas. The other coordinate axis and the directrix are parallel.
Note:
Students must remember the conditions of parabolas, circles, and straight lines to solve these types of questions. Note that while finding the resultant equation which represents the circle, ellipse, or parabola, we need to know the equation of the parabola, circle, ellipse, etc. So that we can easily find the correct answer by analyzing their equations.
Complete step by step Solution:
Given equation is
$\left.\left.\sqrt{(}(x-2)^{2}+y^{2}\right)+\sqrt{(}(x+2)^{2}+y^{2}\right)=4$
$\left.\left.\sqrt{(}(x-2)^{2}+y^{2}\right)=4-\sqrt{(}(x+2)^{2}+y^{2}\right)$
Squaring both sides
$\left.(x-2)^{2}+y^{2}=16+\left((x+2)^{2}+y^{2}\right)-8 \sqrt{(}(x+2)^{2}+y^{2}\right)$
$\left.\left(x^{2}+4-4 x+y^{2}\right)=16+x^{2}+4+4 x+y^{2}-8 \sqrt{(}(x+2)^{2}+y^{2}\right)$
Simplify the terms
$\left.-8 x=16-8 \sqrt{(}(x+2)^{2}+y^{2}\right)$
$\left.x=-2+\sqrt{(}(x+2)^{2}+y^{2}\right)$
$\left.(x+2)=\sqrt{(}(x+2)^{2}+y^{2}\right)$
Squaring again
$(x+2)^{2}=(x+2)^{2}+y^{2}$
$=>y^{2}=0$
$=>y=0$
Here the $y$ axis value is 0
That means there is no line that passes through the $y$ axis
This implies that the line passes through $x$ axis and it will be a straight line.
$=>x$ axis
The location of all the points in a plane whose sum of the distances from two fixed points in the plane is constant is known as an ellipse. The foci (singular focus), which are fixed points and are encircled by the curve, are known. Directrix is the fixed line, and the eccentricity of the ellipse is the constant ratio.
Then the ellipse, circle, or parabola will not be formed by this equation
Hence, the correct option is 2.
Additional information:
We know for a straight line value comes out to be zero.
And A circle is a collection of all points in a plane that is uniformly distanced from a fixed point. The radius of a circle is the separation between the center and any point along its circumference.
we can arrive at the following parabola equations:
$y^{2}=4 a x$
$y^{2}=-4 a x$
$x^{2}=4 a y$
$x^{2}=-4 a y$
The term "standard equations of parabolas" refers to these four equations. It is significant to note that one of the coordinate axes, the vertex at the origin, is the focus of the common equations for parabolas. The other coordinate axis and the directrix are parallel.
Note:
Students must remember the conditions of parabolas, circles, and straight lines to solve these types of questions. Note that while finding the resultant equation which represents the circle, ellipse, or parabola, we need to know the equation of the parabola, circle, ellipse, etc. So that we can easily find the correct answer by analyzing their equations.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series
