Question

$P:2x^2-axy+6y^2=0$
$Q:3x^2-8xy+4y^2=0$

If one of the lines represented by $P$ coincides with one of the lines represented by $Q$ and the other two are perpendicular, then the value of $a$ is

• A. $1$
  
• B. $2$
• C. $-1$
• D. none of these

Answer 1

The correct option is D none of these

$Q:3x^2-8xy+4y^2=0$
$Q:3x^2-6xy-2xy+4y^2=0$
$Q:3x(x-2y)-2y(x-2y)=0$
$Q:(3x-2y)(x-2y)=0$
.
One of the lines of $Q$ is the same as $P$
$P:2x^2-axy+6y^2=0$
Let it be $x-2y=0$
$P:2x^2-axy+6y^2=(x-2y)(kx-my)$
$P:2x^2-axy+6y^2=k{x}^2-(2k+m)xy+2m{y}^2$
Comparing we get,
$k=2,m=3$, which gives $2x-3y=0$, which is not perpendicular to $3x-2y=0$
.
Hence, for the same pair one of the lines be $3x-2y=0$
$P:2x^2-axy+6y^2=(3x-2y)(kx-my)$
$P:2x^2-axy+6y^2=3k{x}^2-(2k+3m)xy+2m{y}^2$
Comparing we get,
$k=\frac{2}{3},m=3$, which gives $2x-9y=0$, which is not perpendicular to $3x-2y=0$

Thus none of them satisfies the equation.