Analytical method to solve the given equations I require the points of intersection of two curves $y^2=4x$ and $y=e^{-x/2}$ to the find the angle between them. Is there any method to find the points as when I tried to graph them the solution was a not so obvious value 
I did get the answer by using the formula as the denominator becomes 0 so the curves are orthogonal but is there some way to solve these 2 equations.
 A: Squaring your second equation we get $$y^2=e^{-x}$$ so you have to solve
$$4x=e^{-x}$$
It is $$\left\{\left\{x\to W\left(\frac{1}{4}\right)\right\}\right\}$$
A: It sounds like you already solved the problem this way, but it's worth repeating that you don't need an exact value of the intersection point to answer the question.
As the good Doctor points out, the curves intersect where $4x=e^{-x}$.  Let $h(x) = 4x-e^{-x}$.  Since $h(0) = -1$ and $h(1) = 4-1/e > 0$, there is a zero of $h$ at some point $x_*$ between $0$ and $1$.  Since $h'(x) = 4+e^{-x} > 0$ for all $x$, there are no other zeroes of $h$ (between two zeroes of $h$ would lie a zero of $h'$). The curves intersect at $(x_*,y_*)$ where $y_*^2 = 4x_*$ and $y_* = e^{-x_*/2}$.
Along the first curve,
$$
    y^2 = 4x \implies 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}
$$
Therefore,
$$
    \left.\frac{dy}{dx}\right|_{(x_*,y_*)} = \frac{2}{y_*}
$$
Along the second curve,
$$
    y = e^{-x/2} \implies \frac{dy}{dx} = -\frac{1}{2} e^{-x/2} = - \frac{1}{2} y
$$
Therefore,
$$
    \left.\frac{dy}{dx}\right|_{(x_*,y_*)} = -\frac{1}{2} y_*
$$
You can see that these two tangent lines have negative reciprocal slopes, meaning they are orthogonal.
