Show that the line `d/a+y/b=1` touches the curve `y=b e^(-x/a)` at the point where it crosses the y-axis.

Explanation

We have the equation of line given by `x/a + y/b=1`, which touches the curve `y=b.e^(-x//a)` at the point, where the curve intersects the axis of Y i.e., x=0
`therefore y=b.e^(-0//a)=b` `[therefore e^(@)=1]`
So, the point of intersection of the curve with Y-axis is (0,b).
Now, slope of the given line at (0,b) is given by
`1/a.1+1/b.(dy)/(dx)=0`
`rArr (dy)/(dx)=-1/a.b`
`rArr (dy)/(dx) = -1/a. b= -b/a= m_(1)` [say]
Also, the slope of the curve at (0,b) is
`(dy)/(dx)= b.e^(-x//a). -1/a`
`(dy)/(dx) = -b/ae^(-x//a)`
`(dy)/(dx)_(0,b) = -b/ae^(-0)= -b/a=m_(2)` [say]
Since, `m_(1)=m_(2)=-b/a`
Hence, the line touches the curve at the point, where the curve intersects the axis of Y.