Question

If the equations $12x^2+7xy-p{y}^2-18x+qy+6=0$ represents a pair of perpendicular straight lines, then

• A. $p=12,q=1$
• B. $p=1,q=12$
• C. $p=-1,q=12$
• D. $p=1,q=-12$

Answer 1

The correct option is A $p=12,q=1$

Given equation is
$12x^2+7xy-p{y}^2-18x+qy+6=0$
On comparing with
$a{x}^2+b{y}^2+2hxy+2gx+2fy+c=0$, we get
$a=12,b=-p,h=\frac{7}{2},g=-9,f=\frac{q}{2},c=6$
Conditions for pair of lines and pair of perpendiculars are
$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$ and $a+b=0$
$\Rightarrow 12\times (-p)\left(6\right)+2\times \left(\frac{q}{2}\right)(-9)\left(\frac{7}{2}\right)-12{\left(\frac{q}{2}\right)}^2-(-p)(-9{)}^2-6{\left(\frac{7}{2}\right)}^2=0$
and $12-p=0$
$\Rightarrow -72p-\frac{63}{2}q-3q^2+81p-\frac{147}{2}=0$
and $12-p=0$
$\Rightarrow 2q^2+21q-23=0$ and $p=12$
$\Rightarrow q=1$ and $p=12$.