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The ellipse x^2+""4y^2=""4 is inscribed ...

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+12y^(2)=16`

B

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

C

`4x^(2) + 64y^(2) = 49`

D

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

Text Solution

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A
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Knowledge Check

  • Ellipse x^(2) + 4y^(2) = 4 is inscribed in a rectangle aligned with co-ordinate axes . This rectangle itself is inscribed in another ellipse the passes ellipse that through (-4,0) . Then the equation of the ellipse is

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

  • The coordinates of a focus of the ellipse 4x^(2) + 9y^(2) =1 are

    A
    `((sqrt(5))/(6),0)`
    B
    `(-(sqrt(5))/(6),0)`
    C
    `((sqrt(3))/(6),0)`
    D
    `(-(sqrt(3))/(6),0)`

  • The distnce between the foci of the ellipse 3x ^(2) + 4y ^(2) =48 is

    A
    2
    B
    4
    C
    6
    D
    8

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