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