A) \[ax+by=1\]
B) \[ax-by=1\]
C) \[\frac{x}{a}-\frac{y}{b}=1\]
D) \[\frac{x}{a}+\frac{y}{b}=1\]
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),\[{{\left( \frac{dy}{dx} \right)}_{(0,\,b)}}=\frac{-b}{a}\] \Equation of tangent is, \[y-b=\frac{-b}{a}(x-0)\] \[\Rightarrow \]\[\frac{x}{a}+\frac{y}{b}=1\].
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