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Q.
$PQ$ and RS are normal chord to parabola $y ^2=8 x$ at $P \& R$ on the curve respectively and the points $P, Q, R, S$ are concyclic. If $A$ is vertex of parabola, then
Conic Sections
Solution:
Equation of normal chords at $P \left(2 t _1{ }^2, 4 t _1\right) \& R \left(2 t _2{ }^2, 4 t _2\right)$ are
$y + t _1 x -4 t _1-2 t _1{ }^3=0 \& y + t _2 x -4 t _2-2 t _2{ }^3=0$
Equation of any two degree curve through $P , Q , R , S$ is
$\left(y+t_1 x-4 t_1-2 t_1^3\right)\left(y+t_2 x-4 t_2-2 t_2^3\right)+\lambda\left(y^2-8 x\right)=0$
$\Theta P, Q, R, S$ are concyclic
$\therefore 1+\lambda= t _1 t _2 \& t _1+ t _2=0$
$\therefore$ centroid of $\triangle APR$ lies on $x$-axis
& slope of $PR =\frac{4 t _1-4 t _2}{2 t _1{ }^2-2 t _2{ }^2}=\frac{2}{ t _1+ t _2}$
$\therefore P R$ is parallel to $y$-axis.