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
$$\eqalign{
& y - 3 = \frac{{3 + 2}}{{0 - 5}}\left( {x - 0} \right),\,{\text{i}}{\text{.e}}{\text{., }}x + y - 3 = 0 \cr
& y = \frac{c}{{x + 1}}{\text{ or }}\frac{{dy}}{{dx}} = \frac{{ - c}}{{{{\left( {x + 1} \right)}^2}}} \cr} $$
Let the line touch the curve at $$\left( {\alpha ,\,\beta } \right).$$
Then $$\alpha + \beta - 3 = 0\,{\left( {\frac{{dy}}{{dx}}} \right)_{\alpha ,\,\beta }} = \frac{{ - c}}{{{{\left( {\alpha + 1} \right)}^2}}} = - 1,$$ and $$\beta = \frac{c}{{\alpha + 1}}$$
$$\eqalign{
& \therefore \,\frac{c}{{\left( {\frac{c}{\beta }} \right)}} = 1 \cr
& {\text{or, }}{\beta ^2} = c \cr
& {\text{or, }}{\left( {3 - \alpha } \right)^2} = c = {\left( {\alpha + 1} \right)^2} \cr
& {\text{or, }}{\left( {3 - \alpha } \right)^2} = {\left( {\alpha + 1} \right)^2} \cr
& {\text{or, }}3 - \alpha = \pm \left( {\alpha + 1} \right) \cr
& {\text{or, 3}} - \alpha = \alpha + 1 \cr
& {\text{or, }}\alpha = 1 \cr
& {\text{So, }}c = {\left( {1 + 1} \right)^2} = 4 \cr} $$