Let P, Q, and R are three co-normal points on the parabola $y^{2} = 4 a x$ . Then correct statement (s) is / are

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algebraic sum of the slopes of the normals at P, Q, and R vanishes.

algebraic sum of the ordinates of the points P, Q, and R vanishes.

centroid of the triangle PQR lies on the axis of the parabola

Circle circumscribing the triangle PQR passes through the vertex of the parabola

answer is A, B, C, D.

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Detailed Solution

Equation of normal in slope form is
$\begin{array}{l} y = m x - 2 a m - a m^{3} \\ a m^{3} + 2 a m - m x + y = 0 \\ a m^{3} + (2 a - x) m + y = 0 \\ l e t P = (a m_{1}^{2} - 2 a m_{1}) Q = (a m_{2}^{2}, - 2 a m_{2}) R = (a m_{3}^{2}, - 2 a m_{3}) \\ m_{1} + m_{2} + m_{3} = 0 \\ \dot{i} . e a \lg e b r a i c s u m o f t h e s l o p e s o f t h e n o r m a l s a t P, Q a n d R v a n i s h e s . \\ P, Q a n d R v a n i s h e s \\ y - c o - o r d i n a t e o f c e n t r o i d o f Δ P Q R i s z e r o \\ ∴ c e n t r o i d l i e s o n x - a x i s \\ c i r c l e c i r c u m s c r i b i n g t h e t r i a n g l e P Q R a l w a y s p a s s e s t h r o u g h t h e v e r t e x o f t h e p a r a b o l a . \end{array}$