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The eccentricity of an ellipse, with its centre at the origin, is 1/2. If one of the directrices is x=4 , then the equation of the ellipse is :

  • Option 1)

    4x^{2}+3y^{2}=12

  • Option 2)

    3x^{2}+4y^{2}=12

  • Option 3)

    3x^{2}+4y^{2}=1

  • Option 4)

    4x^{2}+3y^{2}=1

 

Answers (1)

As we learnt in

Standard equation -

\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1
 

- wherein

a\rightarrow Semi major axis

b\rightarrow Semi minor axis

 

 

 

Eccentricity -

e= \sqrt{1-\frac{b^{2}}{a^{2}}}

- wherein

For the ellipse  

\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1

 

 

 

Equation of directrices -

x= \pm \frac{a}{e}

- wherein

For the ellipse  

\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1

 

 

e=\frac{1}{2} \:and\: \frac{a}{e}=4

 Thus\: \:\frac{a}{\frac{1}{2}}=4

=>a=2

Also \:b^{2}=a^{2}\left ( 1-e^{2} \right )=4\left ( 1-\frac{1}{4} \right )=3

\frac{x^{2}}{4}+\frac{y^{2}}{3}=1

=> 3x^{2}+4y^{2}=12


Option 1)

4x^{2}+3y^{2}=12

Incorrect option    

Option 2)

3x^{2}+4y^{2}=12

Correct option

Option 3)

3x^{2}+4y^{2}=1

Incorrect option    

Option 4)

4x^{2}+3y^{2}=1

Incorrect option    

Posted by

Vakul

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