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Question: The equation of the tangent to the curve \(y = be^{- x/a}\) at the point where it crosses y-axis is...

The equation of the tangent to the curve y=bex/ay = be^{- x/a} at the point where it crosses y-axis is

A

ax+by=1ax + by = 1

B

axby=1ax - by = 1

C

xayb=1\frac{x}{a} - \frac{y}{b} = 1

D

xa+yb=1\frac{x}{a} + \frac{y}{b} = 1

Answer

xa+yb=1\frac{x}{a} + \frac{y}{b} = 1

Explanation

Solution

Curve is y=bex/ay = be^{- x/a}

Since the curve crosses y-axis (i.e., x=0x = 0) \therefore y=by = b

Now dydx=baex/a\frac{dy}{dx} = \frac{- b}{a}e^{- x/a}. At point (0, b), (dydx)(0,b)=ba\left( \frac{dy}{dx} \right)_{(0,b)} = \frac{- b}{a}

\therefore Equation of tangent is yb=ba(x0)y - b = \frac{- b}{a}(x - 0)xa+yb=1\frac{x}{a} + \frac{y}{b} = 1.