The value of $m$,for which the line $y = mx + \frac{25\sqrt{3}}{3}$ is a normal to the conic $\frac{x^2}{16} - \frac{y^2}{9} = 1$,is

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

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