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

Options

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

Answer

B

Explanation

The given ellipse is inscribed in a rectangle PQRS as shown in Fig. 16. Clearly, coordinates of P are (2, 1). The rectangle PQRS is inscribed in another ellipse passing through (4, 0). So, its semi-major axis is 4.

Let the equation of the ellipse be `x^(2)/a^(2) + y^(2)/b^(2) = 1`, where a = 4.
This passes through (2, 1).
`therefore 4/a^(2) + 1/b^(2) = 1 rArr 1/4 + 1/b^(2) = 1 rArr b^(2) = 4/3 " "[because a = 4]`
Hence, the equation of the ellipse is `x^(2)/16 + (3y^(2))/(4) = 1`
or, `x^(2) + 12y^(2) = 16.`