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  • Need clarity, kindly explain! - Co-ordinate geometry - JEE Main-3

The ellipse x^{2}+4y^{2}= 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is

  • Option 1)

    x^{2}+12y^{2}=16

  • Option 2)

    4x^{2}+48y^{2}=48

     

  • Option 3)

    4x^{2}+64y^{2}=48

  • Option 4)

    x^{2}+16y^{2}=16

 

Answers (1)

As we learnt in

Stanard equation -

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

- wherein

a\rightarrow Semi major axis

b\rightarrow Semi minor axis

 Equation is  \frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1

It passes through (2, 1)

Solving, b^{2}=\frac{4}{3}

Hence equation is 

\frac{x^{2}}{16}+\frac{3y^{2}}{4}=1

x^{2}+3y^{2}=16

 


Option 1)

x^{2}+12y^{2}=16

Correct

Option 2)

4x^{2}+48y^{2}=48

 

Incorrect

Option 3)

4x^{2}+64y^{2}=48

Incorrect

Option 4)

x^{2}+16y^{2}=16

Incorrect

Posted by

Vakul

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