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 ]

• A. 6

Answer 1

The correct option is A 6

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 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 $\theta =\frac{P{P}_1}{P1T}$ --------(1)

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.

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

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

This way, we can find the derivative easier

=> f’(x) = $\frac{1}{(1+x{)}^2}$

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

=> f’(x1) = $\frac{1}{(1+x1{)}^2}=\frac{1}{9}$

=> $(1+x1{)}^2=9$

=> x1 = 2 or -4

But x1 > 0 because it is in first quadrant

=> x1 = 2

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

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

= PP1 = ⅔ and tan $\theta =\frac{1}{9}$
=> P1 T = ⅔ 9 = 6