The line joining the points $(0,3)$ and $(5,-2)$ is a tangent to the curve $y=\frac{c}{x+1}$, then $c=$

Slope of line joining the points $(0,3)$ and $(5,-2)$ is $\frac{3+2}{0-5}=-1$ This is equal to the slope of tangent on the curve and that is given by $\frac{d y}{d x}=\frac{-c}{(x+1)^2}$ $\Rightarrow \quad \frac{d y}{d x}=-1$ $\Rightarrow \quad c=(x+1)^2$ ...(i) Equation of line joining the points $(0,3)$ and $(5,2)$ is given by $(y-3)=-1(x-0)$. Solving equation of tangent and the curve for point of intersection $\frac{c}{x+1}+x=3$ ...(ii) Solving Eqs. (i) and (ii), $x=1$ On putting this in Eq. (ii), we get $c=4$