Maximum length of chord of the ellipse $\frac{{{x^2}}}{8} + \frac{{{y^2}}}{4} = 1$, such that eccentric angles of its extremities differ by $\frac{\pi }{2}$ is 

  • A

    $4$

  • B

    $2\sqrt 2 $

  • C

    $16$

  • D

    $8$

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