Question

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

• A. $k=1$
• B. $k=\frac{1}{2}$
• C. $k=2$
• D. $k=-\frac{1}{2}$

Answer 1

The correct option is C

$k=2$

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

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

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

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

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

$\Rightarrow \frac{k}{2}=1\Rightarrow k=2$,
Which is the required condition.