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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.
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.
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