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

$\mathrm{x}=2 \Rightarrow \mathrm{y}=\frac{8}{4+{\mathrm{x}}^2}=\frac{8}{4+4}=1$

$\therefore$ Point (2, 1)  $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-8}{{(4+{\mathrm{x}}^2)}^2}\left(2\mathrm{x}\right)$

Slope Tangent $=\frac{-16\times 2}{64}=-\frac{-1}{2}$

$\therefore$ eq of Tangent is $\mathrm{y}-1=\frac{-1}{2}(\mathrm{x}-2)$

                              $\begin{array}{rcl}2\mathrm{y}-2& =& -\mathrm{x}+2\\ \mathrm{x}+2\mathrm{y}& =& 4\end{array}$

area $=\frac{4^2}{2|1\times 2|}=\frac{16}{4}=4$