Question
If the line joining the points $$\left( {0,\,3} \right)$$ and $$\left( {5,\, - 2} \right)$$ is a tangent to the curve $$y = \frac{c}{{x + 1}}$$ then the value of $$c$$ is :
A.
1
B.
$$-2$$
C.
4
D.
none of these
Answer :
4
Solution :
The equation of the line is $$y - 3 = \frac{{3 + 2}}{{0 - 5}}\left( {x - 0} \right),{\text{i}}{\text{.e}}{\text{., }}x + y - 3 = 0$$
$$y = \frac{c}{{x + 1}}\,\,\,\,\,\,\, \Rightarrow \frac{{dy}}{{dx}} = \frac{{ - c}}{{{{\left( {x + 1} \right)}^2}}}$$
Let the line touches the curve at $$\left( {\alpha ,\,\beta } \right)$$
$$\eqalign{
& \therefore \alpha + \,\beta - 3 = 0, \cr
& {\left. {\,\,\frac{{dy}}{{dx}}} \right)_{\alpha ,\,\beta }} = \frac{{ - c}}{{{{\left( {\alpha + 1} \right)}^2}}} = - 1\,\,\,{\text{and }}\beta = \frac{c}{{\alpha + 1}} \cr
& \therefore \frac{c}{{{{\left( {\frac{c}{\beta }} \right)}^2}}} = 1\,\,{\text{or }}{\beta ^2} = c\,\,{\text{or }}{\left( {3 - \alpha } \right)^2} = c = {\left( {\alpha + 1} \right)^2} \cr
& \therefore 3 - \alpha = \pm \left( {\alpha + 1} \right)\,\,{\text{or }}3 - \alpha = \alpha + 1 \cr
& \therefore \alpha = 1 \cr
& {\text{So, }}c = {\left( {1 + 1} \right)^2} = 4 \cr} $$,