Question

The value of $\lambda$ for which the equation${{x}^{2}}-{{y}^{2}}-x-\lambda y-2=0$ represents a pair of straight lines, are
1. $3,\,\,-3$
2. $-3,\,\,1$
3. $3,\,\,1$
4. $-1,\,\,1$

Answer 1

here in question we have to find the value of $\lambda$ from the equation ${{x}^{2}}-{{y}^{2}}-x-\lambda y-2=0$
Which represents a straight line that means the determinant of the matrix that is $\Delta =0$. The equation is given that we have to write in the form of a matrix then we have to take the determinant and equate it to zero.

Complete step by step answer:
Given equation is that
${{x}^{2}}-{{y}^{2}}-x-\lambda y-2=0$
Now we have to compare with general equation of a circle that is
$a{{x}^{2}}+2hx+b{{y}^{2}}+2gx+2fy+c=0$
Where, $a,\,\,h,\,\,b,\,\,g,\,\,f$ and $\,c$ are constant.
By comparing this we get:
$a=1,\,\,b=-1,\,\,h=0,\,\,g=-\dfrac{1}{2},\,\,f=\dfrac{\lambda }{2},\,\,c=-2$
According to the condition of the straight line, the determinant should be zero.
That is $\Delta =0$
The given general equation can be written in the form of the matrix form to take the determinant of it.
$a{{x}^{2}}+2hx+b{{y}^{2}}+2gx+2fy+c=0$
It can be written in the form of matrix:

a & h & g \\\ h & b & f \\\ g & f & c \\\ \end{matrix} \right]$$ Now we have to take the determinant of this matrix and equate it to 0. $$\left| \begin{matrix} a & h & g \\\ h & b & f \\\ g & f & c \\\ \end{matrix} \right|=0$$ After solving above equation we get: $$a(bc-{{f}^{2}})-h(hc-gf)+g(hf-bg)=0$$ After simplifying and modifying the above equation we get: $$abc-a{{f}^{2}}-{{h}^{2}}c+ghf+ghf-b{{g}^{2}}=0$$ After solving further we get: $$abc-a{{f}^{2}}-{{h}^{2}}c+2ghf-b{{g}^{2}}=0---(1)$$ After substituting the values of $$a=1,\,\,b=-1,\,\,h=0,\,\,g=-\dfrac{1}{2},\,\,f=\dfrac{\lambda }{2},\,\,c=-2$$ in equation$$(1)$$ we get: $$\left( 1\times (-1)\times (-2) \right)-\left( 1\times {{\left( -\dfrac{\lambda }{2} \right)}^{2}} \right)-({{0}^{2}}\times (-2))+2\left( \left( \dfrac{-1}{2} \right)\times 0\times \left( \dfrac{-\lambda }{2} \right) \right)-\left( (-1)\times {{\left( \dfrac{-1}{2} \right)}^{2}} \right)=0$$ After simplifying this equation we get: $$2-\dfrac{{{\lambda }^{2}}}{4}-0+0+\dfrac{1}{4}=0$$ Here, you can see that there is only one constant to find that is $$\lambda $$. By simplifying we get: $$\dfrac{-{{\lambda }^{2}}}{4}+\dfrac{9}{4}=0$$ $$\dfrac{-{{\lambda }^{2}}}{4}=\dfrac{-9}{4}$$ By solving this, $$-{{\lambda }^{2}}=-9$$ $${{\lambda }^{2}}=9$$ So, we get two roots of $$\lambda $$ That is $$\lambda =+3$$ Or $$\lambda =-3$$ **So, the correct answer is “Option 1”.** **Note:** In this particular problem we have to understand the main condition which is that the given equation is a pair of straight lines. That means in that case remember that the determinant is always zero. Keep in mind that before taking the determinant first we have to write in the form of a matrix then it becomes easier to take the determinant. While comparing the values, note the value of $$a,\,\,b,\,\,h,\,\,g,\,\,f,\,\,c$$ correctly. Because sometimes silly mistakes can happen while copying down the values especially for $$h,\,\,g$$ and $$f$$. So, in this way we can solve similar types of problems.