# The eccentricity of an ellipse \[9{x^2} + 16{y^2} = 144\] is

(a) \[\dfrac{{\sqrt 3 }}{5}\]

(b) \[\dfrac{{\sqrt 5 }}{3}\]

(c) \[\dfrac{{\sqrt 7 }}{4}\]

(d) \[\dfrac{2}{5}\]

Answer

Verified

328.5k+ views

Hint: Here the eccentricity of an ellipse is a measure of how nearly circular is the ellipse. Eccentricity is found by the formula eccentricity = c/a where ‘c’ is the distance from the centre to the focus of the ellipse and ‘a’ is the distance from the centre to the vertex for the standard form of the ellipse.

Given ellipse is \[9{x^2} + 16{y^2} = 144\]

Rewriting the ellipse, we get

\[

\dfrac{{9{x^2}}}{{144}} + \dfrac{{16{y^2}}}{{144}} = 1 \\

\\

\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1 \\

\]

For the ellipse of the form \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] the eccentricity is given by \[e = \dfrac{c}{a}\], where \[c = \sqrt {{a^2} - {b^2}} \].

Comparing both the equations we have \[a = 4,{\text{ }}b = 3{\text{ }}\]

So, \[c = \sqrt {16 - 9} = \sqrt 7 \]

Therefore, \[e = \dfrac{c}{a} = \dfrac{{\sqrt 7 }}{4}\]

Thus, the answer is option (c) \[\dfrac{{\sqrt 7 }}{4}\].

Note: The eccentricity of the ellipse is always greater than zero but less than one i.e. \[0 < e < 1\]. The standard form of the ellipse is \[\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1\] with centre \[\left( {h,k} \right)\]. In this problem we have centre \[\left( {0,0} \right)\] so we have used the ellipse form \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\].

Given ellipse is \[9{x^2} + 16{y^2} = 144\]

Rewriting the ellipse, we get

\[

\dfrac{{9{x^2}}}{{144}} + \dfrac{{16{y^2}}}{{144}} = 1 \\

\\

\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1 \\

\]

For the ellipse of the form \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] the eccentricity is given by \[e = \dfrac{c}{a}\], where \[c = \sqrt {{a^2} - {b^2}} \].

Comparing both the equations we have \[a = 4,{\text{ }}b = 3{\text{ }}\]

So, \[c = \sqrt {16 - 9} = \sqrt 7 \]

Therefore, \[e = \dfrac{c}{a} = \dfrac{{\sqrt 7 }}{4}\]

Thus, the answer is option (c) \[\dfrac{{\sqrt 7 }}{4}\].

Note: The eccentricity of the ellipse is always greater than zero but less than one i.e. \[0 < e < 1\]. The standard form of the ellipse is \[\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1\] with centre \[\left( {h,k} \right)\]. In this problem we have centre \[\left( {0,0} \right)\] so we have used the ellipse form \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\].

Last updated date: 28th May 2023

•

Total views: 328.5k

•

Views today: 5.85k

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Name the Largest and the Smallest Cell in the Human Body ?

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main

A sample of an ideal gas is expanded from 1dm3 to 3dm3 class 11 chemistry CBSE