For the function $f(x)=\frac{x^2}{x^2-1}$, the area formed by the tangent to the curve at $P(2,\frac{4}{3})$ with the coordinate axes is:

Slope of tangent $m=f^{\prime }(2)=\frac{-2(2)}{(2^2-1{)}^2}=\frac{-4}{9}$.Equation of tangent at $(2,4/3)$:$y-\frac{4}{3}=-\frac{4}{9}(x-2)$$9y-12=-4x+8⟹4x+9y=20$.$x$ -intercept (put $y=0$): $4x=20⟹x=5$. Point is $(5,0)$.$y$ -intercept (put $x=0$): $9y=20⟹y=20/9$. Point is $(0,20/9)$.Area of triangle $=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 5\times \frac{20}{9}=\frac{50}{9}$ sq. units.