If $2x-y+1=0$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{16}=1$ , then which of the following CANNOT be sides of a right angled triangle?

$\therefore {\left(1\right)}^2=4a^2-16 \Rightarrow a^2=\frac{17}{4}$

$\Rightarrow a=\frac{\sqrt{17}}{2}$

For $2a,\text{ 4, 1}$, sides are $\sqrt{17}, 4, 1$ ($\Rightarrow$ Right angled triangle)

For $\text{2a, 8, 1}$, sides are $\sqrt{17}, 8, 1$ ($\Rightarrow$Triangle is not possible)

For $\text{a, 4, 1}$, sides are $\frac{\sqrt{17}}{2}, 4, 1$ ($\Rightarrow$Triangle is not possible)

For $\text{a, 4, 2}$, sides are $\frac{\sqrt{17}}{2}, 4, 2$ ($\Rightarrow$Triangle exist but not right angled)