Consider a parabola $y^2 = 4ax$ , parameterize it as $x=at^2$ and $y=2at$, then it is found that if we have a line segment passing through focus, with each points having value of $t$ as $t_1$ and $t_2$ for the parameterization, then it must be that:
$$ t_1 \cdot t_2 = -1$$
Hope for hints.
Chord passing through $(at_1^2,2at_1)$ and $(at_2^2,2at_2)$ is $$y-2at_2=\frac{2at_1-2at_2}{at_1^2-at_2^2}(x-at_2^2)$$ $$y-2at_2=\frac{2}{t_1+t_2}(x-at_2^2)$$
It should pass through focus $(a,0)$ $$\Rightarrow -at_2=\frac{1}{t_1+t_2}\,a(1-t_2^2)$$ $$\Rightarrow t_1t_2=-1$$
Hint:
Let the parameters be $p,q$. So the eq of the line is $(y-2ap)(ap^2-aq^2)=(x-ap^2)(2ap-2aq)$. What point does this pass through so that it is a focal chord?
Working back from geometry makes everything supremely trivial, following blue's post, it must be that at the tangent vectors at end point of focal chord is perpendicular.
Our curve is given as $\gamma(y) = ( \frac{y^2}{4a},y)$, the tangent vector is given as: $\tau(y) = (\frac{y}{2a},1)$, let the two points we want be at height $(y_1,y_2)$, then due to tangents at end point being perpendicular:
$$ \tau(y_1) \cdot \tau(y_2)=0$$
Expanding everything out: $$ y_1 y_2 = -4a$$
Now plugging in the parameterization, we find the parametric version of the result.