#### Answer

(a) $y = \frac{1}{4}x +\frac{ln(4)+1}{4}$
(b) $y = e~x$

#### Work Step by Step

(a) $x-4y = 1$
$4y = x-1$
$y = \frac{1}{4}x-\frac{1}{4}$
If the tangent line is parallel to this line, then the slope of the tangent line must have the same slope of $\frac{1}{4}$
We can find the point on the curve $y = e^x$ where the slope is $\frac{1}{4}$:
$y = e^x$
$y' = e^x = \frac{1}{4}$
$ln~(e^x) = ln(\frac{1}{4})$
$x = ln(\frac{1}{4})$
$x = -ln(4)$
When $x = ln(\frac{1}{4}),~~$ then $~~y = e^{ln(1/4)} = \frac{1}{4}$
The tangent line meets the curve $y = e^x$ at the point $(-ln(4), \frac{1}{4})$
We can find the equation of the tangent line:
$y - \frac{1}{4} = \frac{1}{4}[x - (-ln(4)]$
$y = \frac{1}{4}x +\frac{1}{4}~ln(4)+\frac{1}{4}$
$y = \frac{1}{4}x +\frac{ln(4)+1}{4}$
(b) The tangent line meets the curve at the point $(x,e^x)$
If the tangent line passes through the origin, then the slope of the line is $m = \frac{e^x-0}{x-0} = \frac{e^x}{x}$
Also:
$m = y' = e^x$
Then $~~e^x = \frac{e^x}{x},~~$ and so $~~x = 1$
The slope of the tangent line is $m = e^1 = e$
When $x = 1~~$ then $~~y = e$
The tangent meets the curve at the point $(1,e)$
We can find the equation of the tangent line:
$(y-e) = e(x-1)$
$y = e~x-e+e$
$y = e~x$