A) (0, 0)
B) (0, a)
C) (0, b)
D) (b, 0)
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 \[\frac{dy}{dx}=\frac{-b}{a}{{e}^{-x/a}}\] \[{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}=\frac{-b}{a}{{e}^{-{{x}_{1}}/a}}=\frac{-{{y}_{1}}}{a}\] (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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