If the two sides of a parallelogram are represented by $a x^{2} + 2 h x y + b y^{2} = 0 a n d p x + q y = 1$ is one of its diagonals, then the other diagonal is

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$y (b p - h q) = x (a q - h p)$

$x (b p - h q) = y (a q - h p)$

$y (b p + h q) = x (a q + h p)$

$x (b p + h q) = y (a q + h p)$

answer is B.

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Detailed Solution

$\begin{array}{l} L e t a x^{2} + 2 h x y + b y^{2} = 0 r e p r e s e n t t h e l i n e s \\ y = m_{1} x - - (1) a n d y = m_{2} x - - (2) \\ a n d p x + q y - 1 = 0 - - (3) \\ s u b (1) i n (3) \\ \Rightarrow p x + q m_{1} x = 1 \\ \Rightarrow x = \frac{1}{p = q m_{1}}, y = \frac{m_{1}}{p + q m_{1}} \\ ∴ A = (\frac{1}{p + q m_{1}}, \frac{m_{1}}{p + q m_{1}}) \\ S u b (2) i n (3) \\ \Rightarrow p x + q m_{2} x = 1 \\ \Rightarrow x = \frac{1}{p + q m_{2}}, y = \frac{m_{2}}{p + q m_{2}} \\ B (\frac{1}{p + q m_{2}}, \frac{m_{2}}{p + q m_{2}}) \end{array}$

$\begin{array}{l} p (α, β) = m i d p o int o f A B = (\frac{1}{2} (\frac{1}{p + q m_{1}} + \frac{1}{p + q m_{2}}), \frac{1}{2} (\frac{m_{1}}{p + q m_{1}} + \frac{m_{2}}{p + q m_{2}})) \\ m = S l o p e o f O P = \frac{β}{α} \\ = \frac{\frac{m_{1}}{p + q m_{1}} + \frac{m_{2}}{p + q m_{2}}}{\frac{1}{p + q m_{1}} + \frac{1}{p + q m_{2}}} \\ \Rightarrow \frac{p m_{1} + q m_{1} m_{2} + p m_{2} + q m_{1} m_{2}}{p + q m_{2} + p + q m_{1}} \end{array}$

$\begin{array}{l} = \frac{p (m_{1} + m_{2}) + 2 q m_{1} m_{2}}{2 p + q (m_{1} + m_{2})} \\ = \frac{p (\frac{- 2 h}{b}) + 2 q (\frac{a}{b})}{2 p + q (\frac{- 2 h}{b})} \\ = \frac{a q - p h}{p b - q h} \\ E q u a t i o n o f \bar{O C} i s y = m x \\ \Rightarrow y = \frac{a q - p h}{p b - q h} x \\ \Rightarrow y (p b - q h) = (a q - p h) x \end{array}$