Question

The value of A with $|\lambda |<16$ such that $2x^2-10xy+12y^2+5x+\lambda y-3=0$ represents a pair of straight lines, is ?

• A. $-10$
• B. $-9$
• C. $10$
• D. $9$

Answer 1

The correct option is B $-9$

Given

$2x^2-10xy+12y^2+5x+\lambda y-3=0$ is pair of straight line

So $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

$2\times 12\times (-3)+2\frac{\lambda }{2}\times \frac{5}{2}\times (-5)-2(\frac{\lambda }{2}{)}^2-(12){\left(\frac{5}{2}\right)}^2-(-3\left)\right(-5{)}^2=0$

$144+25\lambda +{\lambda }^2=0$

$\lambda =\frac{-25\pm \sqrt{625-576}}{2}$

$\lambda =\frac{-25\pm \sqrt{49}}{2}$

$\lambda =\frac{-25\pm 7}{2}$

$\lambda =-9,-16$

But $\left|\lambda \right|<16$

$\lambda >-16$

Hence $\lambda =-9$