If a normal drawn to the ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$ touches the hyperbola $\frac{x^2}{4} - \frac{y^2}{3} = 1$,then the square of the slope of that normal is

  • A
    $3$
  • B
    $4$
  • C
    $9$
  • D
    $16$

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