Question

Find the length of subtangent on the curve $y=\frac{x}{1+x}$, where the slope of the tangent is $\frac{1}{9}$

[ The point where the tangent is drawn is in first quadrant ]

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Answer 1

We saw that in the given figure, PT is the length of tangent and P1T is the length of subtangent. We are given slope of the tangent, that is $\theta$ as per the figure. In the figure, PP1 is the y-coordinate of the point where the tangent is drawn.

We have slope of tangent =$tan\theta =\frac{PP1}{P1T}-------\left(1\right)$

We will now find the point where the tangent is drawn. We are given the slope is $\frac{1}{9}$ We can use this to find the point where the tangent is drawn.

$y=\frac{x}{1+x}$. We will write this as $y=\frac{x+1-1}{1+x}$

$\Rightarrow y=1-\frac{1}{1+x}$

This way, we can find the derivative easier

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{(1+x{)}^2}$

Let the point where the slope is $\frac{1}{9}$ be x1

$\Rightarrow f^{\prime }\left(x1\right)=\frac{1}{(1+x1{)}^2}=\frac{1}{9}\Rightarrow (1+x1{)}^2=9\Rightarrow x1=2or-4$

But x1>0 because it is in first quadrant

$\Rightarrow x1=2$

Y-coordinate = length of $PP1=\frac{2}{2+1}=\frac{2}{3}$ (substituting x1 =2 in the equation of the curve)

From (1) we get P1T = $\frac{PP1}{tan\theta }$

$=PP1=\frac{2}{3}$ and $tan\theta =\frac{1}{9}$

$\Rightarrow P1T=\frac{2}{3}\times 9=6$