#### 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-ln2.

#### 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=-ln12

Since -lnx=lnx-1

So x=ln2

Put x=ln2 in (1), we get y=e-ln2=1/2.

So, the points are ln2, 12 and the slope of the tangent is -2. Therefore, the equation of tangent is

y-12=-12x-ln2

**Conclusion:**

The equation of the tangent is y-12=-12x-ln2.