To determine

To find:

We have to find an equation of the tangent to the curve

y=e-x (1)

 which is

perpendicular to the line 2x-y=8 (2)

Answer

 The equation of the tangent is y-12=-12x-ln⁡2.

Explanation

Calculation:

Equation of the line is 2x-y=8 and its gradient is m1=dydx.

Differentiate (2) with respect to x, we have

2=dydx

So m1=2.

Since, the equation of the tangent is perpendicular to the line (2). So, if its slope is m2, then

m1.m2=-1

After putting the value of m1, we have m2=-1/2.

Let the point of tangency be the point (x, y) on the curve (1) and the slope of tangent on the curve is obtained by differentiating (1) with respect to x.

So

dydx=-e-x=m2=-1/2

e-x=1/2

Taking natural logarithmic on both sides, we have

x=-ln⁡12

Since -ln⁡x=ln⁡x-1

So x=ln⁡2

Put x=ln⁡2 in (1), we get y=e-ln⁡2=1/2.

So, the points are ln⁡2,  12 and the slope of the tangent is -2. Therefore, the equation of tangent is

y-12=-12x-ln⁡2

Conclusion:

 The equation of the tangent is y-12=-12x-ln⁡2.