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Let F(x) = (1+b^(2))x^(2) + 2bx + 1. The...

Let `F(x) = (1+b^(2))x^(2) + 2bx + 1`. The minimum value of F(x) is the reciprocal of the square of the eccentricity of the hyperbola `(x^(2))/(16) + (y^(2))/(9) =1`. Then the equation of the common tangents to the curves `(x^(2))/(16) + (y^(2))/(9//16)=1` and `x^(2) + y^(2) = b^(2)` is

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