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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. |
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Answer» The equation of the curve is `y = be^(-x//a)` ,rbgt It crosses the y-axis at the point where `x = 0` Putting x = 0 in the equation of the curve, we get `y = b` So, the point of contact is (0, b) Now, `y = be^(-x//a) rArr (dy)/(dx) = (-be^(-x//a))/(a)` `:. ((dy)/(dx))_((0","b)) = - (b)/(a)` So, the equaiton of the tangent is `( y- b)/(x - 0) = (-b)/(a) rArr bx + ay = ab rArr (x)/(a) + (y)/(b) = 1` Hence, `(x)/(a) + (y)/(b) = 1` touches the curve `y = be^(-x//a) " at " (0, b)` |
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