Question

the condition for a conic $x^2+2xy+2y+kx+3y^2=0$ to represent a pair of straight lines

• A. k=1
• B. k=2
• C. k=3
• D. k=4

Answer 1

The correct option is B k=2

This is a straight forward condition. For a conic of the general form,

$a{x}^2+2hxy+2gx+2fy+b{y}^2=0$ to represent a pair of straight line, the required condition in determinant form is,

$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|$ =0

The given conic equation is,

$x^2+2xy+2y+kx+3y^2=0$

Where,

a =1 , b = 1 , h = 1 , g = $\frac{k}{2}$ , f = 1, c = 0;

Putting these values in the condition we get,

$\left|\begin{array}{ccc}1& 1& \frac{k}{2}\\ 1& 1& 1\\ \frac{k}{2}& 1& 1\end{array}\right|$ =0

$\left|\begin{array}{ccc}1& 1& \frac{k}{2}-1\\ 1& 1& 1\\ \frac{k}{2}& 1& 1\end{array}\right|$ =0

$(\frac{k}{2}-1)(1-\frac{k}{2})=0$

$\frac{k}{2}=1$,

k= 2, which is the required condition.