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Answer 1 · 2018-05-02T17:04:59

$c = 61/4$ and $c= -65/4$ .

Normal means perpendicular to the curve. The given normal line can be rewritten as follows:

$y = 1/2x + c/2$

This means that the slope is $1/2$ . The slope of the tangent at that point will be $-1/(1/2) = -2$ .

Thus we must find the values of $x$ where the derivative equals $-2$ .

$y = 4(2x - 1)^-1$

The derivative as given by the chain rule is

$y' = -8/(2x- 1)^2$

Set $y' = -2$ .

$-2 = -8(2x- 1)^2$

$1/4 = (2x- 1)^2$

$+- 1/2 = 2x - 1$

$2x= 3/2 and 2x = 1/2$

$x = 3/4 and x = 1/4$

The curve's corresponding y-values are given by calculating using the function

$y(3/4) = 4/(2(3/4) - 1) = 4/(1/2) = 8$
$y(1/4) = 4/(2(1/4) - 1) = 4/(-1/2) = -8$

We now solve for $C$ knowing that $2y= x + c$ .

$2(8) = 3/4 + c -> c= 16 - 3/4 -> c = 61/4$
$2(-8) = 1/4 + c-> c = -16 - 1/4 -> c = -65/4$

Hopefully this helps!