Question

The values of $\lambda$, for which the equation$x^2-y^2-x+\lambda y-2=0$ represents a pair of straight lines, is

• A. $-3,1$
• B. $-1,1$
• C. $3,-3$
• D. $3,3$

Answer 1

The correct option is C

$3,-3$

Explanation for correct option:

Given $x^2-y^2-x+\lambda y-2=0$

Comparing given equation with the general equation

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

We get $a=1,h=0,b=-1,g=-\frac{1}{2},f=\frac{\lambda }{2},c=-2$

$abc+2fgh–a{f}^2–b{g}^2–c{h}^2=0$represents a pair of straight lines.

$$\begin{gathered} \therefore 2+0–\frac{{\lambda }^2}{4}+\frac{1}{4}–0=0 \\ \Rightarrow \frac{{\lambda }^2}{4}=\frac{9}{4} \\ \Rightarrow {\lambda }^2=9 \\ \Rightarrow \lambda =\pm 3 \\ \Rightarrow \lambda =3,-3 \end{gathered}$$

Hence, option C is correct.