Question

Tangent to $\frac{x^2}{a^2}-\frac{{\displaystyle y^2}}{{\displaystyle 4^2}}$=1 is

y =mx $\pm \sqrt{a^2{m}^2-16}$

Comparing with y = 2x + 1

m = 2

$\Rightarrow$4a² - 16 = 1

a² =  $\frac{17}{4}$

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

Only 2a, 4, 1 are sides of a right-angled triangle