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The line joining the points $(0,3)$ and $(5,-2)$ is a tangent to the curve $y=\frac{c}{x+1}$, then $c=$
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Verified Answer
The correct answer is:
$4$
Slope of line joining the points $(0,3)$ and $(5,-2)$ is $\frac{3+2}{0-5}=-1$
This is equal to the slope of tangent on the curve and that is given by
$\frac{d y}{d x}=\frac{-c}{(x+1)^2}$
$\Rightarrow \quad \frac{d y}{d x}=-1$
$\Rightarrow \quad c=(x+1)^2$ ...(i)
Equation of line joining the points $(0,3)$
and $(5,2)$ is given by $(y-3)=-1(x-0)$.
Solving equation of tangent and the curve for point of intersection
$\frac{c}{x+1}+x=3$ ...(ii)
Solving Eqs. (i) and (ii),
$x=1$
On putting this in Eq. (ii), we get $c=4$
This is equal to the slope of tangent on the curve and that is given by
$\frac{d y}{d x}=\frac{-c}{(x+1)^2}$
$\Rightarrow \quad \frac{d y}{d x}=-1$
$\Rightarrow \quad c=(x+1)^2$ ...(i)
Equation of line joining the points $(0,3)$
and $(5,2)$ is given by $(y-3)=-1(x-0)$.
Solving equation of tangent and the curve for point of intersection
$\frac{c}{x+1}+x=3$ ...(ii)
Solving Eqs. (i) and (ii),
$x=1$
On putting this in Eq. (ii), we get $c=4$
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