Question 1151169
<br>
An arbitrary point on the graph of the function is (a,1/(a-4)).<br>
We need to have the slope of the line containing (0,0) and (a,1/(a-4) equal to the slope of the graph of y=1/(x-4) at x=a -- that is, equal to the derivative of the function at x=a.<br>
{{{y = 1/(x-4) = (x-4)^(-1)}}}
{{{dy/dx = -(x-4)^(-2) = -1/(x-4)^2}}}<br>
The derivative at x=a is<br>
{{{(-1)/(a-4)^2}}}<br>
The slope of the line containing (0,0) and (a,1/(a-4) is<br>
{{{(1/(a-4))/a = 1/(a^2-4a)}}}<br>
So<br>
{{{1/(a^2-4a) = (-1)/(a-4)^2}}}
{{{(a-4)^2 = -1(a^2-4a)}}}
{{{a^2-8a+16 = -a^2+4a}}}
{{{2a^2-12a+16 = 0}}}
{{{a^2-6a+8 = 0}}}
{{{(a-2)(a-4) = 0}}}
{{{a = 2}}} or {{{a = 4}}}<br>
The function is undefined at x=4; so our solution should be at x=2.<br>
For x=2, the point on the graph is (2,1/(2-4)) = (2,-1/2).<br>
The line through (0,0) and (2,-1/2) -- and therefore the tangent line we are looking for -- is y = (-1/4)x.<br>
A graph of the function and the tangent line through the origin:<br>
{{{graph(400,400,-2,5,-1,1,1/(x-4),-(1/4)x)}}}<br>