Question

If $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ represents a pair of parallel lines, then $\sqrt{\frac{g^2-ac}{f^2-bc}}=$

• A. $\frac{a}{b}$
• B. $\sqrt{\frac{a}{b}}$
• C. $\sqrt{\frac{b}{a}}$
• D. $\frac{b}{a}$

Answer 1

The correct option is B $\sqrt{\frac{a}{b}}$

Let $lx+my+n=0$ and $lx+my+k=0$ be two parallel lines.

Then, their joint equation will be:

$(lx+my+n)(lx+my+k)=0$

or, $l^2{x}^2+m^2{y}^2+2lmxy+(n+k)(lx+my)+nk=0$

Now, comparing with given equation:

$a=l^2;b=m^2;n=lm$

and, $g=\frac{(n+k)l}{2};f=\frac{(n+k)m}{2};c=nk$

Now, $\frac{l^2}{m^2}=\frac{a}{b}$, (as $m_1{m}_2=\frac{a}{b}$).........(1)

So, $\sqrt{\frac{g^2-ac}{f^2-bc}}=\sqrt{\frac{(n+k{)}^2{l}^2-4l^{2}nk}{(n+k{)}^2{m}^2-4m^{2}nk}}=\sqrt{\frac{l^2}{m^2}}$$=\sqrt{\frac{a}{b}}$ [using eqn (1)]