Question

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 (1) `x^2+""16 y^2=""16` (2) `x^2+""12 y^2=""16` (3) `4x^2+""48 y^2=""48` (4) `4x^2+""64 y^2=""48`
A. `x^(2)+16y^(2)=16`
B. `x^(2)+12y^(2)=16`
C. `4x^(2)+48y^(2)=16`
D. `4x^(2)+64y^(2)=48`

Answer 1

`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`