If e^x is a function where e is the euler's constant, then from it's property, we know that the slope of e^x at any point (x,y) is e^x itself. Meaning, by looking at the graph, if you draw a tangent at any point, the tan of angle made by that tangent is equal to the Y co-ordinate of the point. Draw a perpendicular from the tangent on to the x-axis to touch it at a point P(x,0) and also extend the tangent to meet the x-axis at a point Q(a,0).
the tan of angle is equal to the Y-co-ordinate of the chosen point on the curve divided by distance between the perpendicular point P(x,0) and the point Q(a,0), which is equal to x-a. Let's say that x-a is the "intercept" made by the tangent on the x-axis, then as the slope equal the y-coordinate,
tan(theta) = y = e^x {theta = angle made by the tangent with the x-axis.)..... (1)
Now, tan(theta) = y/(x-a) = e^x/intercept.....(2)
From the above two equations, intercept = 1.
How can we define the exponential function from the above case?
I mean to ask, can we derive the formula for e^x using this property?
I believe that you have got your formulae slightly wrong, but I believe I understand what you are asking. The Image below shows the exponential function in red, with tangent line at $x=1$ in blue, and vertical line at $x=1$ in green. Now the $\theta$ in your question is the angle that the blue tangent line makes with the $x$-axis at the origin.
We in fact see that the $\tan(\theta)$ is the gradient of the line, and also the ratio of the vertical length of green line to the Horizontal distance between the green line and the blue tangent on the x-axis. -Not to do with the perpendicular- if I have understood correctly. To answer your question it is in fact possible to define the exponential function in this way...
The requirements we have here are, a differentiable function $f(x)$, such that at each $a\in\Bbb{R}$ we have the following property of the tangent line to $f(x)$ at $a$, which is: $$(y-f(a)) = f'(a)(x-a)$$
And we require the distance between the $x$-intercept of this tangent line to be $1$ unit from $x=a$. So when plugging in $y=0$ in to the formula for the tangent line we require an $x$-value of $a-1$.
All this gives: $$(0-f(a)) = f'(a)((a-1)-a)$$
Which is equivalent to $f(a) = f'(a)$ and we require this to be true at any $a\in\Bbb{R}$
This is a separable differential equation so not too hard to solve:
$\frac{f'(x)}{f(x)} = 1\quad$ so $\ln(f(x)) = x+c$
and $f(x) = Ae^x$ and we in fact can see that your desired property gives the family of exponential functions $Ae^x$ with$A\in\Bbb{R}$