We have to find an equation of the tangent to the curve
perpendicular to the line 2x-y=8 (2)
The equation of the tangent is y-12=-12x-ln2.
Equation of the line is 2x-y=8 and its gradient is m1=dydx.
Differentiate (2) with respect to x, we have
Since, the equation of the tangent is perpendicular to the line (2). So, if its slope is m2, then
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.
Taking natural logarithmic on both sides, we have
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