Question 89

# If $$x^{2} — 4y^{2} — x + \lambda y — 2 = 0$$ is to represent a pair of straight lines, then the product of the possible values of $$\lambda$$ is

Solution

Suppose we have two lines $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2=0$$
The equation of pair of these two equations would be obtained by multiplying these two equations,
Giving us, $$a_1a_2x^2+b_1b_2y^2+x\left(a_1c_2+a_2c_1\right)+y\left(b_1c_2+b_2c_1\right)+xy\left(a_1b_2+a_2b_1\right)+c_1c_2=0$$
Comparing this with the equation we have, we can start finding the values,
$$a_1=a_2=1$$
From coefficient of x, we get $$c_1+c_2=-1$$
From the constant in the equation we get, $$c_1c_2=-2$$
The value of $$c_{1\ }and\ c_2$$ in some order must be -2 and 1

From coefficient of xy, we get $$b_2=-b_1$$
And through coefficient of $$y^2$$, we get that $$b_{1\ }or\ b_2=\pm\ 2$$
Since the coefficients of x are equal in both the equations, b becomes the only differentiating point.
The equations will therefore now be,
$$x+2y+c_1=0$$ and $$x-2y+c_2=0$$
Where we can have two case:
Case 1: $$c_1=-2\ \&\ c_2=1$$
The required value would be : -6

Case 2: $$c_2=-2\ \&\ c_1=1$$
The required value would be : 6

The product of possible values therefore is -36