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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.

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2 Answers 2

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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.

enter image description here

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  • $\begingroup$ I'll attempt it $\endgroup$
    – Guest Man
    Commented May 16, 2017 at 1:27
  • $\begingroup$ @Guest Man There must be two tangent lines then. $\endgroup$ Commented May 16, 2017 at 1:29
  • $\begingroup$ If you draw a tangent to all the point (just imagine it), only at one point the tangent is perpendicular to the given line: imgur.com/a/KPrxR $\endgroup$
    – Guest Man
    Commented May 16, 2017 at 1:33
  • $\begingroup$ @Guest Man You graph the derivative and not the original. $\endgroup$ Commented May 16, 2017 at 1:48
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The line

$x - 2y - 1 = 0 \tag{1}$

may also be written

$y = \dfrac{1}{2}x - \dfrac{1}{2}; \tag{2}$

from this we see its slope is $\frac{1}{2}$. Thus the slope of a line normal or perpendicular to this line is $-2$.

For any value of $x$ other than $\frac{1}{2}$, the slope of the curve

$y(x)= \dfrac{1}{2x - 1} = (2x - 1)^{-1} \tag {3}$

is given by

$y'(x) = -2(2x - 1)^{-2}, \tag{4}$

We seek the points on the curve (3) where the slope is $-2$, yielding

$-2(2x - 1)^{-2} = -2, \tag{5}$

or

$(2x - 1)^{-2} = 1, \tag{6}$

or

$(2x - 1)^2 = 1; \tag{7}$

thus

$2x - 1 = \pm 1, \tag{8}$

whence

$x = 0, 1. \tag{9}$

That

$y'(0) = -2, y'(1) = -2 \tag{10}$

is easily checked using (4); thus the slope of the curve (3) is normal to the line (2) at the points $(0, -1)$ and $(1, 1)$.

Note added in edit: it will be observed that the value of $b$, that is, the $y$-intercept of the tangent lines to (3), does not enter in to the above calculations, by reason of the fact that the problem does not call for these lines, only for certain points on the given curve. However, having found these points where the slope is $-2$, the corresponding $b$-values are easily discovered. Each of these lines is of slope $-2$, and is represented by an equation of the form

$y = -2x + b; \tag{11}$

if $(0, -1)$ lies on such a line, we must have

$-1 = -2 \cdot 0 + b, \tag{12}$

or

$b = -1; \tag{13}$

the other $b$-value satisfies

$1 = -2 \cdot 1 + b, \tag{14}$

whence

$b = 3. \tag{15}$

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