Question

If $p,q>0$ and $p<q$ and $p{x}^2+4\mu xy+q{y}^2+4a(x+y+1)=0$ represents a pair of straight lines, then

• A. $p\le a\le q$
• B. $a\le p$
• C. $a\ge p$
• D. $a\le p$ or $a\ge q$

Answer 1

The correct option is D $a\le p$ or $a\ge q$

Given equation is

$p{x}^2+4\mu x^2+q{y}^2+4ax+4ay+4a=0$ ...(1)

Since equation (1) represents a pair of straight lines,

$\therefore △=0$

$\Rightarrow p.q.4a+2.2a.2a.2\mu -p.4a^2-q.4a^2-4a.4{\mu }^2=0$

$\Rightarrow 16a{\mu }^2-16a^{2}u+4a[(p+q)a-pq]=0$

Since $\mu$ is real,$b^2-4ac\ge 0$

$\Rightarrow 256a^4-256a^2[(p+q)a-pq]=0$

$\Rightarrow a^2-(p+q)a+pq\ge 0$

$\therefore (a-p)(a-q)\ge 0$