How to derive the formula to determine the length of chord intercepted by a parabola on a line? The formula to determine the length of chord intercepted by a parabola $y^2=4ax$ on a line $y=mx+c$, where the variables have their usual meaning is given by the following:
$$L=\left(\frac 4 {m^2}\right)\sqrt{a(1+m^2)(a-mc)}$$
The above formula was given in my book without any proof. I searched the internet to my best, and could not even find this formula, and so its derivation. I tried to derive this formula to find whether it works for all cases. I tried using equation of chord derived from the parametric form of the coordinates, but it gets even more complicated.
So, How to derive this formula? 
 A: The given formula can be easily derived by following the steps mentioned below.
Substituting $y=mx+c$ in $y^2=4ax$, we get,
$$m^2x^2+(2mc-4a)x+c^2=0\tag1$$
Similarly, let us obtain a quadratic in $y$ by substituting $x=\frac{y-c}{m}$ in $y^2=4ax$.
$$my^2-4ay+4ac=0\tag2$$
Algebraically, for the above equations to have two distinct roots or, geometrically, for the line to intersect the parabola at two distinct points, the discriminant of the above equations must be greater than zero. Considering equation $(1)$ or $(2)$, the discriminant is $16a^2-16amc$. So the condition $a(a-mc)>0$ must be satisfied. Otherwise, the length of the chord will either be zero or not defined depending upon the case. Let us assume this condition is satisfied and proceed further.
The following graph depicts the given situation:

The line intersects the parabola at two points $A$ and $B$. Let the coordinates of $A$ and $B$ be $(x_1,y_1)$ and $(x_2,y_2)$ respectively. Let the length of the chord $AB$ be $L$, so by distance formula:
$$L=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$
$$L=\sqrt{(x_1+x_2)^2-4x_1x_2+(y_1+y_2)^2-4y_1y_2}\tag3$$
From $(1)$, we can say, $x_1+x_2=\frac{4a-2mc}{m^2}$ and $x_1x_2=\frac{c^2}{m^2}$. Similarly from $(2)$, $y_1+y_2=\frac{4a}m$ and $y_1y_2=\frac{4ac}{m}$. Now, let us plug these values in $(3)$ to get the following:
$$L=\sqrt{\left(\frac{4a-2mc}{m^2}\right)^2-4\left(\frac{c^2}{m^2}\right)+\left(\frac{4a}m\right)^2-4\left(\frac{4ac}{m}\right)}$$
On simplification, we would finally arrive at the following equation:
$$L=\left(\frac 4 {m^2}\right)\sqrt{a(1+m^2)(a-mc)}$$
Which is the one given we are supposed to prove. Hence proved! 
A: Substituting $y=mx+c$ in $y^2=4ax$ gives
$$m^2x^2+(2mc-4a)x+c^2=0$$
Now, using $$y_1-y_2=m(x_1-x_2),\quad x_1+x_2=\frac{4a-2mc}{m^2},\quad x_1x_2=\frac{c^2}{m^2}$$
where $(x_1,y_1),(x_2,y_2)$ are the intersection points, we get
$$\begin{align}L&=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
\\\\&=\sqrt{(x_1-x_2)^2+m^2(x_1-x_2)^2}
\\\\&=\sqrt{(1+m^2)(x_1-x_2)^2}
\\\\&=\sqrt{(1+m^2)((x_1+x_2)^2-4x_1x_2)}
\\\\&=\sqrt{(1+m^2)\left(\left(\frac{4a-2mc}{m^2}\right)^2-4\cdot \frac{c^2}{m^2}\right)}
\\\\&=\sqrt{(1+m^2)\cdot \frac{16a^2-16amc+4m^2c^2-4m^2c^2}{m^4}}
\\\\&=\frac{4\sqrt{a(1+m^2)(a-mc)}}{m^2}\end{align}$$
