Applications Of Derivatives Question 147
Question: The equation of the tangent to curve $ y=b{e^{-x/a}} $ at the point where it crosses y-axis is
[Karnataka CET 2002]
Options:
A) $ ax+by=1 $
B) $ ax-by=1 $
C) $ \frac{x}{a}-\frac{y}{b}=1 $
D) $ \frac{x}{a}+\frac{y}{b}=1 $
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Answer:
Correct Answer: D
Solution:
Curve is $ y=b{e^{-x/a}} $
Since the curve crosses y-axis (i.e., x = 0) \ $ y=b $
Now $ \frac{dy}{dx}=\frac{-b}{a}{e^{-x/a}} $ . At point (0, b), $ {{( \frac{dy}{dx} )} _{(0,,b)}}=\frac{-b}{a} $ \
Equation of tangent is, $ y-b=\frac{-b}{a}(x-0) $
$ \Rightarrow $ $ \frac{x}{a}+\frac{y}{b}=1 $ .
BETA
