Show that the line x/a+y/b=1, touches the curve y=b×e^-x/a at the point where the curve intersect the axis of y.

Question

Arun · Answer

The curve cuts the y-axis where x=0, so at the point P(0,b).
The tangent to the curve has slope
dy/dx=-(b/a)e^(-x/a) =-(b/a) at P.
The equation of this tangent is (y-b)=(-b/a)x
which , dividing by b gives y/b - 1 = (-1/a)x
giving x/a+y/b=1 and this is the given line.
So the result has been shown.

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

Show that the line x/a+y/b=1, touches the curve y=b×e^-x/a at the point where the curve intersect the axis of y.

Show that the line x/a+y/b=1, touches the curve y=b×e^-x/a at the point where the curve intersect the axis of y.

1 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Differential Calculus

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship