Find the sum of the squares of the reciprocals of the two parts of the focal chord of a parabola.
My attempt:
Let $y^2=4ax$ be a parabola. Let PQ be the focal chord through the focus S$(a, 0)$ of the parabola such that the co-ordinates of P & Q are $(at_1^2, 2at_1)$ & $(at_2^2, 2at_2)$. Then we have, $t_1t_2=-1$. Clearly, if $l=PS$ and $l'=QS$, then $l=\sqrt{(at_1^2-a)^2+(2at_1-0)^2}=\dots=a(t_1^2+1)$ and similarly, $l'=a(t_2^2+1)=a\frac{t_1^2}{1+t_1^2}$ (since, $t_1t_2=-1$).
Then $\frac{1}{l}+\frac{1}{l'}=\dots=\frac{1}{a}$.
Then $\frac{1}{l^2}+\frac{1}{l'^2}=\frac{1}{a}-2\frac{1}{ll'}$
But the answer should be independent of $l$ & $l'$. What to do?
Loosely speaking, the "axis chord" has parts $a$ and $\infty$, so the sum of the squares of the reciprocals of the lengths is $$\frac{1}{a^2}+\frac{1}{\infty^2} = \frac{1}{a^2} + 0 = \frac{1}{a^2}$$ However, the focal chord perpendicular to the axis has parts $2a$ and $2a$, so the corresponding sum is $$\frac{1}{4a^2} + \frac{1}{4a^2} = \frac{1}{2a^2}$$ The sum is not independent of the choice of chord.
As you have shown, though: The sum of the (un-squared) reciprocals of the lengths is independent of that choice.