If 2 x - y + 1 = 0 is a tangent to the hyperbola x 2 a 2 - y 2 16 = 1 , then which of the following CANNOT…

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

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

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)