The area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

  • A
    $\sqrt{ab}$
  • B
    $\frac{a}{b}$
  • C
    $2ab$
  • D
    $ab$

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Let $C$ be the circle of minimum area enclosing the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $e = \frac{1}{2}$ and foci $(\pm 2, 0)$. Let $PQR$ be a variable triangle,whose vertex $P$ is on the circle $C$ and the side $QR$ of length $2$ is parallel to the major axis of $E$ and contains the point of intersection of $E$ with the negative $y$-axis. Then the maximum area of the triangle $PQR$ is:

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