Area of the triangle formed by the tangent at  $x=2$ on the curve $y=\frac{8}{4+x^2}$with the coordinate axes is ( in square units)

$\frac{dy}{dx}=\frac{-8\times 2\times \left(2x\right)}{{\left(4+x^2\right)}^2}$

$\text{ Slope }\left(m\right)=\frac{-16\times 4}{(16{)}^2}=\frac{-1}{4}$

$\text{ If }x=2\text{ then }y=\frac{1}{2}$

$\text{ Equation of tangent is }y-\frac{1}{2}=\frac{-1}{4}(x-2)$

$\begin{array}{r}2y-1=\frac{-1}{2}(x-2)\\ x+4y-4=0\end{array}$

$\text{ Area }\left(A\right)=\frac{16}{2\times 1\times 4}=2$