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Question

Find the length of subtangent on the curve y=x1+x, where the slope of the tangent is 19

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


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Solution

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 θ 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 =tanθ=PP1P1T(1)

We will now find the point where the tangent is drawn. We are given the slope is 19 We can use this to find the point where the tangent is drawn.

y=x1+x. We will write this as y=x+111+x

y=111+x

This way, we can find the derivative easier

f(x)=1(1+x)2

Let the point where the slope is 19 be x1

f(x1)=1(1+x1)2=19(1+x1)2=9x1=2or4

But x1>0 because it is in first quadrant

x1=2

Y-coordinate = length of PP1=22+1=23 (substituting x1 =2 in the equation of the curve)

From (1) we get P1T = PP1tanθ

=PP1=23 and tanθ=19

P1T=23×9=6


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