
Find the eccentricity of the conic \[\left[ {\dfrac{{{{(x + 2)}^2}}}{7}} \right] + {\left( {y - 1} \right)^2} = 14\] .
A. \[\sqrt {\dfrac{7}{8}} \]
B. \[\sqrt {\dfrac{6}{{17}}} \]
C. \[\sqrt {\dfrac{3}{2}} \]
D. \[\sqrt {\dfrac{6}{{11}}} \]
E. \[\sqrt {\dfrac{6}{7}} \]
Answer
163.8k+ views
Hint: The equation provided in the question can be transformed into an equation of ellipse from which we will get \[a\] and \[b\] . Then the value of \[a\]and \[b\] we shall put it in the formula of eccentricity.
Formula Used:
The formula of eccentricity is \[\sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \] , where a and b are the constants of the general equation of the ellipse.
Complete step by step solution:
We have to write the equation given in the question in the form of the general equation of ellipse which is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] .
Simplifying the equation into the general equation we will get –
\[\left[ {\dfrac{{{{(x + 2)}^2}}}{7}} \right] + {\left( {y - 1} \right)^2} = 14\]
\[\dfrac{{{{\left( {x + 2} \right)}^2}}}{{{{\left( {\sqrt {98} } \right)}^2}}} + \dfrac{{{{\left( {y - 1} \right)}^2}}}{{{{\left( {\sqrt {14} } \right)}^2}}} = 1\]
\[\dfrac{{{{\left( {x + 2} \right)}^2}}}{{{{\left( {7\sqrt 2 } \right)}^2}}} + \dfrac{{{{\left( {y - 1} \right)}^2}}}{{{{\left( {\sqrt {14} } \right)}^2}}} = 1\] ------ (i)
This is the equation of ellipse from which we will get the values of \[a\] and \[b\] which we will be substituting in the formula of eccentricity.
Now from the equation (i) we will get the values of \[a\] and \[b\] as –
\[a = 7\sqrt 2 \] and \[b = \sqrt {14} \] .
Putting the values of \[a\] and \[b\] in the formula we will get –
Eccentricity =\[\sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \]
\[ = \sqrt {\dfrac{{{{\left( {7\sqrt 2 } \right)}^2} - {{\left( {\sqrt {14} } \right)}^2}}}{{{{\left( {7\sqrt 2 } \right)}^2}}}} \]
\[ = \sqrt {\dfrac{{98 - 14}}{{98}}} \]
\[ = \sqrt {\dfrac{{84}}{{98}}} \]
Dividing by 2 from numerator and denominator, we get
\[ = \sqrt {\dfrac{{42}}{{49}}} \]
Dividing by 2 from numerator and denominator, we get
\[ = \sqrt {\dfrac{6}{7}} \]
Hence the answer is (E) which is \[ = \sqrt {\dfrac{6}{7}} \]
Note: Students often confuse with formula of eccentricity of a hyperbola and ellipse. The formula of eccentricity of ellipse is \[\sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \] .
Formula Used:
The formula of eccentricity is \[\sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \] , where a and b are the constants of the general equation of the ellipse.
Complete step by step solution:
We have to write the equation given in the question in the form of the general equation of ellipse which is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] .
Simplifying the equation into the general equation we will get –
\[\left[ {\dfrac{{{{(x + 2)}^2}}}{7}} \right] + {\left( {y - 1} \right)^2} = 14\]
\[\dfrac{{{{\left( {x + 2} \right)}^2}}}{{{{\left( {\sqrt {98} } \right)}^2}}} + \dfrac{{{{\left( {y - 1} \right)}^2}}}{{{{\left( {\sqrt {14} } \right)}^2}}} = 1\]
\[\dfrac{{{{\left( {x + 2} \right)}^2}}}{{{{\left( {7\sqrt 2 } \right)}^2}}} + \dfrac{{{{\left( {y - 1} \right)}^2}}}{{{{\left( {\sqrt {14} } \right)}^2}}} = 1\] ------ (i)
This is the equation of ellipse from which we will get the values of \[a\] and \[b\] which we will be substituting in the formula of eccentricity.
Now from the equation (i) we will get the values of \[a\] and \[b\] as –
\[a = 7\sqrt 2 \] and \[b = \sqrt {14} \] .
Putting the values of \[a\] and \[b\] in the formula we will get –
Eccentricity =\[\sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \]
\[ = \sqrt {\dfrac{{{{\left( {7\sqrt 2 } \right)}^2} - {{\left( {\sqrt {14} } \right)}^2}}}{{{{\left( {7\sqrt 2 } \right)}^2}}}} \]
\[ = \sqrt {\dfrac{{98 - 14}}{{98}}} \]
\[ = \sqrt {\dfrac{{84}}{{98}}} \]
Dividing by 2 from numerator and denominator, we get
\[ = \sqrt {\dfrac{{42}}{{49}}} \]
Dividing by 2 from numerator and denominator, we get
\[ = \sqrt {\dfrac{6}{7}} \]
Hence the answer is (E) which is \[ = \sqrt {\dfrac{6}{7}} \]
Note: Students often confuse with formula of eccentricity of a hyperbola and ellipse. The formula of eccentricity of ellipse is \[\sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \] .
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

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

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

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
