Question

What is the value of \[\lambda \] for which the equation \[{x^2} - {y^2} - x + \lambda y - 2 = 0\], represents a pair of straight lines?
A. -3, 1
B. -1, 1
C. 3, -3
D. 3, 1

Answer 1

Hint: First we will compare the given equation to general equation of conic section. Then we will apply the condition of the pair of equations.

Formula used:
The general equation of a conic section is given by
\[a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\]
And this equation represents a pair of straight line if
\[abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0\]
And, if
\[\begin{array}{l}ab - {h^2} = 0\,\,\,{\text{then it is a straight line}}\\ab - {h^2} < 0\,\,\,{\text{then it is a pair of straight lines}}\\ab - {h^2} > 0\,\,\,{\text{then it is a point}}\end{array}\]

Complete step-by-step solution:
Here, we have the equation
\[{x^2} - {y^2} - x + \lambda y - 2 = 0\]
Comparing with the general equation of the conic sections we have,
\[a = 1,b = - 1,c = - 2,h = 0,g = - \dfrac{1}{2},f = \dfrac{\lambda }{2}\]
Substituting these values into the condition expression,
\[abc + 2fgh - a{f^2} - b{g^2} - c{h^2}\]
\[ = 1 \cdot ( - 1) \cdot ( - 2) + 2\dfrac{\lambda }{2} \cdot \dfrac{{ - 1}}{2} \cdot 0 - 1{\left( {\dfrac{\lambda }{2}} \right)^2} - ( - 1){\left( {\dfrac{{ - 1}}{2}} \right)^2} - ( - 2){(0)^2}\]
\[ = 2 + 0 - \dfrac{{{\lambda ^2}}}{4} + \dfrac{1}{4} - 0\]
\[\begin{array}{l} = \dfrac{8 - {\lambda ^2} + 1}{4}\\ = \dfrac{9 - {\lambda ^2}}{4}\end{array}\]
For the equation to be of a straight line this expression must be equal to 0.
\[\begin{array}{l} \dfrac{9 - {\lambda ^2}}{4} = 0\\ \Rightarrow {\lambda ^2} = 9\\ \Rightarrow \lambda = \pm 3\end{array}\]
While the other condition is also satisfied
\[\begin{array}{l}ab - {h^2}\\ \Rightarrow 1( - 1) - 0 = - 1 < 0\end{array}\]
Hence, this equation is of a pair of straight lines if \[\lambda = 3, - 3\].

Therefore, the correct option is option (C).

Note: The formula \[abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0\] can be written as \[\left| {\begin{array}{*{20}{c}}a&h&g\\h&b&f\\g&f&c\end{array}} \right| = 0\]. \[\left| {\begin{array}{*{20}{c}}a&h&g\\h&b&f\\g&f&c\end{array}} \right| = 0\] is easy to remember.