The Tangent to the curve y=xlnx at the point (e,e) meets the x-axis at A and the y-axis at B. Find the distance AB.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You must find the equation of the line AB and then you can find the distance AB. You have a point (e,e) and you must find the slope to write the equation in the point slope form.

To find the slope, you must find the first derivative of the function y.

y'=(xlnx)'

Use the product rule: (ab)'=a'b+ab'

y'=x'lnx+x(lnx)'

y'=lnx+x/x

y'=lnx+1

y(e)=lne+1=1+1=2

The slope is m=2

The equation of AB is: y-e=2(x-e)

y-e-2x+2e=0

y-2x+e=0 => y=2x+e

The segment AB intersects x in A => y=0

0=2x+e =>2x=-e =>x=-e/2

The coordinates for A(-e/2,0)

The segment AB intersects y in B => x=0 => y=e

The coordinates for B(0,e)

Use the distance formula to find AB:

AB = square root [(xB-xA)^2+(yB-yA)^2]

AB = square root (e^2/4 + e^2)

AB = square root 5e^2/4

AB = e*square root (5)/2

ANSWER: The distance AB = e*square root (5)/2.

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