Question

The condition that one of the straight lines given by the equation $a{x}^2+2hx+b{y}^2=0$ may coincide with one of those given by the equation $a^{\prime }{x}^2+2h^{\prime }xy+b^{\prime }{y}^2=0$ is

• A. $(a{b}^{\prime }-a^{\prime }b{)}^2=4(h{a}^{\prime }-h^{\prime }a\left)\right(b{h}^{\prime }-b^{\prime }h)$
• B. $(a{b}^{\prime }-a^{\prime }b{)}^2=(h{a}^{\prime }-h^{\prime }a\left)\right(b{h}^{\prime }-b^{\prime }h)$
• C. $(b{a}^{\prime }-b^{\prime }a{)}^2=4(a{b}^{\prime }-a^{\prime }b\left)\right(b{h}^{\prime }-b^{\prime }h)$
• D. $(b{h}^{\prime }-b^{\prime }h{)}^2=4(a{b}^{\prime }-a^{\prime }b\left)\right(h{a}^{\prime }-h^{\prime }a)$

Answer 1

Answer: (A)

$(a{b}^{\prime }-a^{\prime }b{)}^2=4(h{a}^{\prime }-h^{\prime }a\left)\right(b{h}^{\prime }-b^{\prime }h)$

Let the common line be $y=mx$. Then it must satisfy both the equations. Therefore, we have
$b{m}^2+2hm+a=0$ ........ (i)
$b^{\prime }{m}^2+2h^{\prime }m+a^{\prime }=0$ ........ (ii)
Solving Equations (i) and (ii), we gbet
$\frac{m^2}{2(h{a}^{\prime }-h^{\prime }a)}=\frac{m}{a{b}^{\prime }-a^{\prime }b}=\frac{a}{2(b{h}^{\prime }-b^{\prime }h)}$
Eliminating m, we get
${\left[\frac{a{b}^{\prime }-a^{\prime }b}{2(b{h}^{\prime }-b^{\prime }h)}\right]}^2=\frac{h{a}^{\prime }-h^{\prime }a}{b{h}^{\prime }-b^{\prime }h}$
$\Rightarrow (a{b}^{\prime }-a^{\prime }b{)}^2=4(h{a}^{\prime }-h^{\prime }a\left)\right(b{h}^{\prime }-b^{\prime }h)$