Question

If the line joining the points $(0,3)$ and $(5,-2)$ is a tangent to the curve $y=\frac{c}{x+1}$, then find the value of $\frac{c}{8}$ ?

Answer 1

Answer: 0.5

The equation of the line is
$y-3=\frac{3+2}{0-5}(x-0)$,
i.e., $x+y-3=0$
$y=\frac{c}{x+1}$
$\Rightarrow \frac{dy}{dx}=\frac{-c}{(x+1{)}^2}$
Let the line touches the curve at $(\alpha ,\beta )$.
$\Rightarrow \alpha +\beta -3=0,{\left(\frac{dy}{dx}\right)}_{\alpha ,\beta }=\frac{-c}{(\alpha +1{)}^2}=-1$
and $\beta =\frac{c}{\alpha +1}$
$\Rightarrow \frac{c}{(c/\beta {)}^2}=1$
or ${\beta }^2=c$
or $(3-\alpha {)}^2=c=(\alpha +1{)}^2$
$\Rightarrow 3-\alpha =\pm (\alpha +1)$
or $3-\alpha =\alpha +1$
$\Rightarrow \alpha =1.$
So, $c=(1+1{)}^2=4$
$\Rightarrow \frac{c}{8}=0.5$