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The ellipse x^(2)4y^(2)=4 is inscribed i...

The ellipse `x^(2)4y^(2)=4` is inscribed in a rectangle aligned with the coordinates axes, whicj in turn is inscribed in another ellipse that passes through the point (0,0). Then, the equation of the ellipse is

A

`x^(2)+16y^(2)=16`

B

`x^(2)+12y^(2)=16`

C

`4x^(2)+48y^(2)=16`

D

`4x^(2)+64y^(2)=48`

Text Solution

Verified by Experts


`x^(2)+4y^(2)=4rArr(x^(2))/(4)+(y^(2))/(1)=1`
So, `a=2,b=1`
Thus P is (2,1)
The required ellipse is `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 rArr(x^(2))/(4^(2))+(y^(2))/(b^(2))=1`
The point (2,1) lis on it.So
`(4)/(16)+(1)/(b^(2))=1rArr(1)/(b^(2)=1-(1)/(4)=(3)/(4)rArr b^(2)=(4)/(3)`
`:. (x^(2))/(16)+(y^(2))/(((4)/(3)))=1rArr(x^(2))/(16)+(3y^(2))/(4)=1rArr+12y^(2)=16`
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