A tangent to $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ is

$$y = mx + \sqrt {9{m^2} - 4} ,\,m > 0$$ .... (1)

A tangent to $${(x - 4)^2} + {y^2} = 16$$ is

$$xy = m(x - 4) + 4\sqrt {1 + {m^2}} $$ ..... (2)

Comparing (1) and (2),

$$\sqrt {9{m^2} - 4} = - 4m + 4\sqrt {1 + {m^2}} \Rightarrow \sqrt {9 - {4 \over {{m^2}}}} = - 4 + 4\sqrt {1 + {1 \over {{m^2}}}} $$

Let $${1 \over {{m^2}}} = t$$, we have $$\sqrt {9 - 4t} = - 4 + 4\sqrt {1 + t} $$

Squaring, we have

$$ \Rightarrow 9 - 4t = 16 + 16(1 + t) - 32\sqrt {1 + t} \Rightarrow 32\sqrt {1 + t} = 23 + 20t$$

Again squaring $$1024(1 + t) = 529 + 920t + 400{t^2}$$

$$ \Rightarrow 400{t^2} - 104t - 495 = 0 \Rightarrow t = {5 \over 4}$$

Thus $${m^2} = {4 \over 5},\,m = {2 \over {\sqrt 5 }}$$

The tangent is $$y = {2 \over {\sqrt 5 }}x + {4 \over {\sqrt 5 }}$$ i.e. $$2x - \sqrt 5 y + 4 = 0$$