Finding a equation for a tangent line to the curve $y=e^x$ which also goes through the origin I have an assignment

Find an equation for a tangent line to the curve $y=e^x$ which also goes through the origin.

However, in my formula it is asserted 
that the slope between at a point "p" and the origin is 
$m=\frac{e^p-0}{p-o}$ = $\frac{d}{dx}e^x$ (at $x=p$) = $e^p$
but surely $\frac{e^p-0}{p-o} = \frac{e^p}{p}$ how is this equal to 
$\frac{d}{dx}e^x$ (at $x=p$) = $e^p$?
Am I missing something obvious?
Thank you in advance
 A: If you mean the tangent line to the graph then let $(a,e^a)$ be a common point.
Thus, $$y-e^a=e^a(x-a)$$ and since our tangent line goes through the origin,
we obtain
$$0-e^a=e^a(0-a),$$ which gives $a=1$ and the answer: $y=ex$.
A: The troublesome equal sign isn't a conclusion, it's a condition.  The slope of the line is expressed in two different ways.  $m = \frac{e^p}{p}$ by algebra methods and $m=e^p$ by calculus methods.  They are set equal to each other so that you can solve for $p$:
$$\frac{e^p}{p} = e^p$$
$$\frac{1}{p} = 1$$
$$p=1$$
So the slope of the tangent line you're seeking is $e^1$.  Its $y$-intercept is $0$, so the equation is $y=ex$.
A: Problem:

Find an equation for a tangent line to the curve $y=e^{x}$ which also goes through the origin.

Here's how I would solve it:
For a line to be tangent to a given curve, the line and the curve must share a common point. Let's call it $(x_{0},e^{x_{0}})$. Then, the slope of the line tangent to the curve $y=e^{x}$ at $x_{0}$ must be: $\frac{d}{dx}e^{x}|_{x=x_{0}}=e^{x_{0}}$. From that we can fine the line equation of the sought tangent line in terms of the point $x_{0}$:
$$
y-e^{x_{0}}=e^{x_{0}}(x-x_{0})\implies
y=e^{x_{0}}x-e^{x_{0}}(x_{0}-1)
$$
According to the problem statement, the line should also go through the origin:
$$
0=e^{x_{0}}\cdot 0-e^{x_{0}}(x_{0}-1)\implies x_{0}=1\\
$$
Now, just plug in $x_{0}=1$ into our original equation:
$$
y=e^{1}x-e^{1}(1-1)\implies y=ex
$$
Answer: $y=ex$.
