Find the point on the graph of $y=e^{2x}$ at which the tangent line passes through the origin Find the point on the graph of $y=e^{2x}$ at which the tangent line passes through the origin.
Completely lost on this question, the wording is confusing here.
 A: Some Hints:
You know that a line has the slope $y=mx+b$.  To pass through the origin, $b=0$, leaving us with the equation $y=mx$.  The derivative of a function gives you the slope of the tangent line, or in the notation I've been using thus far, $m$.  
So, this question is asking for a point $(x_0, y_0)$ such that $y_0 = mx_0$.
(Hint: $m$ will actually be a function of $x$ and $y$)
If you need more help, just leave a comment.
A: If $(x_0;y_0)$ is your point, then the straight line passing thru it and tangent to the graph is $$y-e^{2x_0}=2e^{2x_0}(x-x_0).$$ If the origin lies on this line, then $x_0=1/2$.
A: Look at the graph of $y = e^{2x}$. For every point on that graph, you can imagine a straight-line that is tangent to it (i.e. goes through that point and has the same slope as the graph at that point). The question's asking you to find the line that goes through the origin; $(0,0)$.
A: We have a mystery point $P$ on the curve. The tangent line at $P$ goes through the origin. We want to find the coordinates of $P$.
Let the coordinates of $P$ be $(a,e^{2a})$.  We find the equation of the tangent line to $y=e^{2x}$ at $P$.
The slope of the tangent line at $P$ is $2e^{2a}$. So the equation of the tangent line is
$$y-e^{2a} =2e^{2a}(x-a).$$
This line passes through the origin. So $(0,0)$ is on the line, and therefore
$$-e^{2a}=(2e^{2a})(-a).$$
Cancel the $e^{2a}$. We get $a=\frac{1}{2}$.
