Applications-Of-Derivatives Question 450
Question: At which point the line $ \frac{x}{a}+\frac{y}{b}=1 $ , touches the curve $ y=b{e^{-x/a}} $
[RPET 1999]
Options:
A) (0, 0)
B) (0, a)
C) (0, b)
D) (b, 0)
Show Answer
Answer:
Correct Answer: C
Solution:
Let the point be $ (x_1,,y_1) $ ,
$ \therefore $ $ y_1=b{e^{-x_1/a}} $ …..(i) Also, curve $ y=b{e^{-x/a}} $
Þ $ \frac{dy}{dx}=\frac{-b}{a}{e^{-x/a}} $ $ {{( \frac{dy}{dx} )}_{(x_1,,y_1)}}=\frac{-b}{a}{e^{-x_1/a}}=\frac{-y_1}{a} $ (by (i)) Now, the equation of tangent of given curve at point $ (x_1,,y_1) $ is $ y-y_1=\frac{-y_1}{a}(x-x_1) $
Þ $ \frac{x}{a}+\frac{y}{y_1}=\frac{x_1}{a}+1 $ Comparing with $ \frac{x}{a}+\frac{y}{b}=1, $ we get $ y_1=b,\text{ and }1+\frac{x_1}{a}=1\Rightarrow x_1=0 $ . Hence, the point is (0, b).
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