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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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