How do I find the point on the curve $y=1/(2x−1)$ where the tangent line will be perpendicular to the line defined by $x−2y−1=0$?
What I have so far:
The equation of the given line is $y= 0.5x+1$
Since the slope is $-0.5$, the tangent slope must be $2$ .: $y=-2x+b$ (tangent line)
The differential of the curve is $y=-2/((2x-1)^2)$.
I can't figure out how to get the b value for the tangent, nor the point at which this happens.
Best Answer
The slope of the tangent line is $$m=y'=\frac{-2}{(2x-1)^2}$$ and since it is perpendicular to the line $x-2y-1=0$ whose slope is $1/2$ we get $m=-2$ and hence, $$-2=\frac{-2}{(2x-1)^2}$$ then solve for $x$. You must get $x=0$ or $x=1$. The points of tangency are $(0,-1)$ and $(1,1)$. The two tangent lines must be $2x+y+1=0$ and $2x+y-3=0$. See the graph below.