Let $A_1(x_1, y_1)$ and $A_2(x_2, y_2)$ be the endpoints of the parabola chord. Lines through the origin i points $A_1$ i $A_2$ are perpendicular. Their slopes are negative reciprocals, so the equation $x_1 x_2 + y_1 y_2 = 0$ stands. Line equation for the parabola chord is of the form $y = k x + l$. If we plug in that $y$ into the parabola equation we will get the quadratic equation which is satisfied by $x_1$ i $x_2$ (abscissa of the intersection of the line and the parabola): $$k^2x^2 + (2kl-4a)x + l^2 = 0$$. Viète's formulas give us $x_1 x_2 = \frac{l^2}{k^2}$. Now, we can calculate $y_1 y_2 = \frac{4al}{k}$. By adding $x_1 x_2$ and $y_1 y_2$ and arranging expressions $l = -4ak$ . We conclude that every parabola chord which is the hypotenuse of a right triangle, with the right angle vertex being at the origin, lies on the line with an equation of the form $y=k(x-4a)$, and those lines obviously pass through the point $(4a, 0)$
Basically, I think I understand everything until after the Viete formulas. Then I lose track at $y_1y_2=4al/k$. I don't follow where does that come from nor how l was derived.
Plugging $x_1,x_2$ we get $y_1,y_2$
$y_1 = kx_1 + l\\ y_2 = kx_2 + l$
Multiplying
$y_1y_2 = (kx_1 + l)(kx_2+l) = k^2x_1x_2 + kl(x_1+x_2) + l_2$
And from Vieta's formulas
$x_1x_2 = \frac {l^2}{k^2}\\ x_1+x_2 = -\frac {2kl-4a}{k^2}$
Putting it together
$y_1y_2 = l^2 + \frac {(4al - 2kl^2)}{k} + l_2 = \frac {4al}{k}$