$I f A, B, C a r e 3 p o i n t s i n a p a r a b o l a . △_{1}, △_{2} a r e t h e a r e a s o f t r i a n g l e i s f o r m e d b y t h e p o i n t s A, B, C a n d t h e \tan g e n t s a t A, B, C . I f △_{1}, △_{2} a r e t h e r o o t s o f p x^{2} + q x + r = 0 t h e n c o n d i t i o n i s$

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$9 q^{2} = 2 pr$

$9 pr = 2 q^{2}$

$9 p^{2} = 2 qr$

$2 p^{2} = 9 qr$

answer is A.

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Detailed Solution

$∆_{1} = \frac{1}{8 a} |(y_{1} - y_{1}) (y_{2} - y_{3}) (y_{3} - y_{1})| ∆_{2} = \frac{1}{16 a} |(y_{1} - y_{2}) (y_{2} - y_{3}) (y_{3} - y_{1})| \frac{∆_{1}}{∆_{2}} = 2 \Rightarrow \frac{∆_{1} + ∆_{2}}{∆_{1} - ∆_{2}} = 3 \Rightarrow ∆_{1} + ∆_{2} = 3 (∆_{1} - ∆_{2}) (\frac{- q}{p}) = 3 \frac{(\sqrt{q^{2} - 4 p r})}{p} \Rightarrow q^{2} = 9 q^{2} - 36 p r \Rightarrow 8 q^{2} = 36 p r \Rightarrow 2 q^{2} = 9 p r$