Problem: The value of k for which the equation $\large 2x^2 + 5y^2 -2kxy + 4x + 6y = 0 $ represent the pair of straight lines is

(A) $ -\frac{19}{6}$

(B) $ \frac{19}{6}$

(C) $ -\frac{7}{6}$

(D) none of these

Ans: (A)

Sol: $\large h^2 \ge ab$ ⇒ $\large k^2 \ge 10 $

and , $\large \left| \begin{array}{ccc} a & h & g \\ h & b & f \\ g & f & c \end{array} \right| = 0$

$\large \left| \begin{array}{ccc} 2 & -k & 2 \\ -k & 5 & 3 \\ 2 & 3 & 0 \end{array} \right| = 0$

⇒ 2(–3k – 10) – 3(6 + 2k) = 0

⇒ –6k – 20 – 18 – 6k = 0

⇒ 12k = –38

k = -38/12 = -19/6

k^{2} = 361/36 > 10. Hence k = –19/6