If the pairs of lines $3 x^{2} - 5 x y + p y^{2} = 0 a n d 6 x^{2} - x y - 5 y^{2} = 0$ have one line in common, then P=

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2, 25/4

-2, 25/4

-2, -25/4

2, -25/4

answer is D.

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Detailed Solution

Given Equation $3 x^{2} - 5 x y + p y^{2} = 0$

$\begin{array}{l} a_{1} = 3, 2 h_{1} = - 5, b_{1} = p \\ a n d 6 x^{2} - x y - 5 y^{2} = 0 \\ a_{2} = 6, 2 h_{2} = - 1, b_{2} = - 5 \end{array}$

The lines have one line in common then

$\begin{array}{l} |\begin{matrix} a_{1} & 2 h_{1} \\ a_{2} & 2 h_{2} \end{matrix}| |\begin{matrix} 2 h_{1} & b_{1} \\ 2 h_{2} & b_{2} \end{matrix}| = {|\begin{matrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{matrix}|}^{2} \\ \Rightarrow |\begin{matrix} 3 & - 5 \\ 6 & - 1 \end{matrix}| |\begin{matrix} - 5 & p \\ - 1 & - 5 \end{matrix}| = |\begin{matrix} 3 & p \\ 6 & - 5 \end{matrix}| \end{array}$

$\begin{array}{l} \Rightarrow (- 3 + 30) (25 + p) = {(- 15 - 6 p)}^{2} \\ \Rightarrow 27 (25 + p) = 9 {(5 + 2 p)}^{2} \\ \Rightarrow 3 (25 + p) = {(5 + 2 p)}^{2} \\ \Rightarrow 75 + 3 p = 25 + 4 p^{2} + 10 p \end{array}$

$\begin{array}{l} \Rightarrow 4 p^{2} + 7 p - 50 = 0 \\ \Rightarrow 4 p^{2} + 25 p - 8 p - 50 = 0 \\ \Rightarrow p (4 p + 25) - 2 (4 p + 25) = 0 \\ \Rightarrow (4 p + 25) (p - 2) = 0 \\ \Rightarrow 4 p + 25 = 0, p - 2 = 0 \\ 4 p = - 25 p = 2 \\ p = - \frac{25}{4} \end{array}$