Line y = mx + C is tangent to hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ if following condition is satisfied:

$${C^2} = {a^2}{m^2} - {b^2}$$ ..... (1)

Given : The equation of line = $$2x - y + 1 = 0$$

Hyperbola = $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{{16}^2}}} = 1$$

Consider equation $$2x - y + 1 = 0$$

$$ \Rightarrow y = 2x + 1 \Rightarrow m = 2$$ and C = 1

From equation of hyperbola, we get b² = 16 and a²

Substituting all values in Eq. (1), we get

$${1^2} = {a^2}{(2)^2} - 16$$

$$1 = 4{a^2} - 16 \Rightarrow 4{a^2} = 16 + 1 \Rightarrow 4{a^2} = 17$$

$$ \Rightarrow {a^2} = {{17} \over 4} \Rightarrow a = \sqrt {{{17} \over 4}} = {{\sqrt {17} } \over 2}$$

Therefore, $$a = {{\sqrt {17} } \over 2}$$

Option (A) : a, 4, 1 $$ \Rightarrow {{\sqrt {17} } \over 2}$$, 4, 1 is not a right triangle (Since $$\sqrt {{4^2} + {1^1}} \ne {{\sqrt {17} } \over 2}$$)

Option (B) : 2a, 4, 1 $$\sqrt {17} $$, 4, 1 is a right angled triangle (since $$\sqrt {{4^2} + {1^2}} = \sqrt {16 + 1} = \sqrt {17} $$)

Option (C) : a, 4, 2 $$ \Rightarrow {{\sqrt {17} } \over 2}$$, 4, 2 is not a right angled triangle (Since $$\sqrt {{4^2} + {2^2}} \ne {{\sqrt {17} } \over 2}$$)

Option (D) : 2a, 8, 1 $$\sqrt {17} $$, 8, 1 is not a right angled triangle (Since $$\sqrt {{8^2} + {1^2}} \ne \sqrt {17} $$)