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 $c$.

Answer 1

line joining $(0,3)(5,-2)$
$y-3=\frac{-2-3}{5-0}(x-0)$
$y-3=-x$
$y=3-x$
touches $y =\frac{ c }{ x +1}$
$(3-x)=\frac{c}{x+1}$
$3 x+3-x^{2}-x=c$
$x^{2}-2 x+(c-3)=0$
$D=0$
$(-2)^{2}-4 \times 1(c-3)=0$
$4-4 c+12=0$
$c=4$