Question

If the equations $x^2+2xy+p{y}^2=0$ and $p{x}^2+2xy+y^2=0$ have one factor exactly in common, then the joint equation of their other two factors will be given by
(correct answer + 1, wrong answer - 0.25)

• A. $3x^2+8xy-3y^2=0$
• B. $3x^2+10xy+3y^2=0$
• C. $y^2+2xy-3x^2=0$
• D. $x^2+2xy-3y^2=0$

Answer 1

The correct option is B $3x^2+10xy+3y^2=0$

Let $y=mx$ be a common factor, then $p{m}^2+2m+1=0$ and $m^2+2m+p=0$
$\Rightarrow \frac{m^2}{2(1-p)}=\frac{m}{p^2-1}=\frac{1}{2(1-p)}$
$\Rightarrow m^2=1$ and $m=-\frac{p+1}{2}$
$\Rightarrow (p+1{)}^2=4$
$\Rightarrow p=1$ or $p=-3$
But for $p=1$, the two pairs have both the lines in common.
So, $p=-3$

Now, $x^2+2xy+p{y}^2=x^2+2xy-3y^2=(x-y)(x+3y)$
and $p{x}^2+2xy+y^2=-3x^2+2xy+y^2=-(x-y)(3x+y)$

$\therefore$ Slope $m$ of the line common to both the pairs is $1$.
So, joint equation of the required lines is $(x+3y)(3x+y)=0$
$\Rightarrow 3x^2+10xy+3y^2=0$