Question

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

Answer 1

Answer:

Please find the attached image.

Step-by-step explanation:

Answer 2

Given : $y = be^\frac{-x}{a}$

Step-by-step explanation:

since , it crosses y axis

then

$y = b e^0 = b$

point where it meets y axis is

(0,b)

now , differentiating with respect to x we get

$\frac{dy}{dx}= \frac{-b}{a}e^\frac{-x}{a}\\\\\left ( \frac{dy}{dx} \right )(_0_,_b)= \frac{-b}{a}e^-^0= \frac{-b}{a}$

then

the equation of the tangent at the point (0,b) is

$y-b = \frac{-b}{a}(x-0)\\\\ay - ab = -bx\\\\\frac{x}{a}+\frac{y}{b}=1$

hence this is the  required equation

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