For a given parabola  $y^2=4ax$, two variable chords PQ and RS at right angles are drawn through the fixed point $A(x_1,y_1)$ inside the parabola, making variable angles 
‘ $\theta$’ and ‘ $\alpha$’ with x-axis respectively. If $r_1,r_2,r_3,r_4$ are distances of P,Q,R and S respectively from A, then the value of  $\frac{1}{r_1{r}_2}+\frac{1}{r_3{r}_4}$.

Let two perpendicular chords through $A\left(x_1{y}_1\right)$ be PQ and RS
Equation of PQ is  $\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }=r$
Where $\tan \theta =$ slope of PQ
Any point on this line may be taken as
 $(x_1+r\cos \theta ,y_1+r\sin \theta )$
As the point lies on $y^2=4ax$ ,
 $$\begin{gathered} \therefore {(y_1+r\sin \theta )}^2=4a(x_1+r\cos \theta ) \\ \Rightarrow r^2{\sin }^2\theta +2y_1\sin \theta r-4a\cos \theta r+y_1^2-4a{x}_1=0 \\ \therefore r_1{r}_2=\frac{(y_1^2-4a{x}_1)}{{\sin }^2\theta } \end{gathered}$$
 Similarly $r_3{r}_4=\frac{(y_1^2-4a{x}_1)}{{\sin }^2\alpha }$ where  $\tan \alpha =$slope of RS
Since PQ is perpendicular to Rs,
 $\therefore \alpha =90+\theta or\theta -90$
In either case;  ${\sin }^2\alpha ={\cos }^2\theta$
Now $\frac{1}{r_1{r}_2}+\frac{1}{r_3{r}_4}=\frac{{\sin }^2\theta +{\cos }^2\theta }{(y_1^2-4a{x}_1)}=\frac{1}{(y_1^2-4a{x}_1)}$ which is constant.