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

Question

Aicharya · Accepted Answer

Equation of line joining $(0,3)$ and $(5,-2)$ is $\begin{aligned} & \frac{y-3}{x}=-1 \Rightarrow x+y-3=0 ...(i)\\ & \because y=\frac{c}{x+1} \Rightarrow \frac{d y}{d x}=\frac{-c}{(x+1)^2}=-1 \\ & c=(x+1)^2 \\ & y=x+1 \Rightarrow x-y+1=0...(ii) \end{aligned}$
Solving (i) and (ii), $x=1, y=2$
$c=(x+2)^2=4$

The value of $c$ such that the straight line joining the points $(0,3)$ and $(5,-2)$ is tangent to the curve…

Solution