Applications Of Derivatives Question 2
Question: The equation of the tangent to the curve $ y=b{e^{-x/a}} $ at the point where it crosses the y-axis is
Options:
A) $ \frac{x}{a}-\frac{y}{b}=1 $
B) $ ax+by=1 $
C) $ ax-by=1 $
D) $ \frac{x}{a}+\frac{y}{b}=1 $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ y=b{e^{-x/a}} $ meets the y-axis at (0, b).
Again, $ y=b{e^{-x/a}}( -\frac{1}{a} ) $
$ At(0,b),\frac{dy}{dx}=be^{0}( -\frac{1}{a} )=-\frac{b}{a} $
Therefore, required tangent is $ y-b=-\frac{b}{a}(x-0) $ or $ \frac{x}{a}+\frac{y}{b}=1 $
BETA
