If the curves $\frac{x^2}{\alpha} + \frac{y^2}{4} = 1$ and $y^3 = 16x$ intersect at right angles,then a value of $\alpha$ is

  • A
    $2$
  • B
    $\frac{4}{3}$
  • C
    $\frac{1}{2}$
  • D
    $\frac{3}{4}$

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